Meet Your Tutor
Electrostatic potential and capacitance become intuitive when every formula is tied to work, energy or charge storage. I will help you track reference points, signs, combinations and dielectrics with a unit and limiting-case check after each numerical.
Take a breath. This chapter has a reputation for being scary, and almost all of that reputation comes from one thing: it uses a lot of symbols. But underneath the symbols there are really only two ideas. The first is potential — a way of asking “how much energy would a charge gain or lose if I put it here?” The second is capacitance — a way of asking “how much charge can this arrangement of conductors hold for every volt I push on it?” Everything else in the chapter is those two ideas dressed up in different clothes.
If you have already worked through Electric Charges and Fields, you are in a very good position. There you learned about force and field — the vector side of electrostatics. Here you learn about energy and potential — the scalar side. Scalars are friendlier. No components, no angles between vectors, no resolving into x and y. You just add numbers, keeping track of plus and minus signs. Many problems that were painful with field vectors become almost easy with potential.
We will go slowly. Every formula gets explained before it gets used, every worked example is done in full with units written at each step, and nothing is assumed. Read a section, close your eyes and try to say it back in your own words, then do the examples with a pen. That is the whole method.
- Electric Potential and Potential Difference
- Potential Due to a Point Charge
- Potential Due to a System of Charges
- Potential Due to an Electric Dipole
- Equipotential Surfaces
- Relation Between Field and Potential
- Potential Energy of a System of Two Point Charges
- Potential Energy of a Dipole in a Uniform Field
- Conductors in an Electrostatic Field: Free and Bound Charges
- Dielectrics and Electric Polarisation
- Capacitors and Capacitance
- The Parallel Plate Capacitor
- Effect of a Dielectric on Capacitance
- Capacitors in Series and in Parallel
- Energy Stored in a Capacitor
- Formula Sheet for Revision
Your Game Plan
Do not try to swallow this chapter in one sitting. Here is the order that works, tested on a lot of students:
- Days 1–2: get comfortable with potential itself. Sections 1 to 4. Learn what V means physically before you touch a single capacitor. Practise adding potentials with signs.
- Day 3: the geometry of potential. Equipotential surfaces and the E–V relation. These two sections are pure understanding, very few numbers, and they show up as 1- and 2-mark questions almost every year.
- Day 4: energy. Potential energy of charge systems and of a dipole. Sign conventions matter more than arithmetic here, so go slowly.
- Day 5: conductors and dielectrics. Mostly theory, mostly free marks if you can state things cleanly.
- Days 6–7: capacitors. Definition, parallel plate, dielectric effect, series and parallel, energy. This is where the long numerical questions live, so give it two full days.
- Day 8: the worksheet at the bottom of this page. Attempt every question with the answers hidden. Only then open the reveals.
Every numerical answer on this page was computed with those exact values. In the board exam you may be given k = 9 × 109, which changes answers only in the third significant figure — that is fine, and examiners accept it.
Study Notes
Electric Potential and Potential Difference
Start with something you already believe. If you lift a stone, you do work against gravity, and that work is stored — let go and the stone comes back down, giving the energy back. Gravity is a conservative force, so the work you do depends only on where the stone started and where it finished, not on the path.
The electrostatic force behaves exactly the same way. Push a positive charge towards another positive charge and you do work against the repulsion; that work is stored as electrostatic potential energy. Because the force is conservative, only the start point and the end point matter.
Now here is the clever move. The stored energy depends on how big the charge you moved was. A 2 μC charge stores twice as much as a 1 μC charge at the same spot. So physicists divide the energy by the charge, and what is left is a property of the location alone. That is electric potential.
V = W∞→P / q and the potential difference is VB − VA = WA→B / q
SI unit: the volt, 1 V = 1 J C−1. Potential is a scalar — it has a sign but no direction.
Two things to notice in that definition. First, we chose infinity as the zero of potential, because at infinity a charge feels nothing. Second, “moved very slowly” means the external agent exactly balances the electric force at every instant, so no energy sneaks off into kinetic energy. That is why the work done becomes pure potential energy.
Why potential difference is what really matters. The zero of potential is our choice, just like the zero of height on a hill. What no one can argue about is the difference between two points, because that difference is a real amount of energy per coulomb. Every battery you have ever used is labelled with a potential difference, never with a potential.
Rearranging the definition gives the single most useful equation in the whole chapter:
W = q × ΔV
Work equals charge times potential difference. Learn it in that form and half the numerical questions in this chapter answer themselves.
Step 1. Convert to SI. q = 2.0 μC = 2.0 × 10−6 C.
Step 2. Apply W = q ΔV = (2.0 × 10−6 C)(12 V).
Step 3. W = 2.4 × 10−5 J.
Unit check: C × V = C × (J/C) = J. Correct. The answer is positive, which tells you the agent had to push — B is at a higher potential and the charge is positive, so it did not want to go.
Step 1. q = 5.0 nC = 5.0 × 10−9 C.
Step 2. VB − VA = W / q = (3.0 × 10−7 J) / (5.0 × 10−9 C).
Step 3. Divide the numbers: 3.0/5.0 = 0.60. Divide the powers: 10−7/10−9 = 102. So 0.60 × 102 = 60 V.
B is 60 volts above A. Notice how the powers of ten did all the real work — get into the habit of handling them separately and you will stop making sign-of-the-exponent slips.
A unit worth loving: the electron volt. When a particle of charge e falls through a potential difference of 1 volt, it gains energy e × 1 V = 1.602 × 10−19 J. We call that one electron volt (eV). It is not an SI unit, but it makes atomic-scale numbers readable.
Step 1 — energy. A charge of magnitude e falling through 200 V gains 200 eV. In joules:
KE = (1.602 × 10−19 C)(200 V) = 3.204 × 10−17 J.
Step 2 — speed. All of it is kinetic energy, so ½mv2 = 3.204 × 10−17 J.
v2 = 2(3.204 × 10−17) / (9.11 × 10−31) = 7.033 × 1013 m2 s−2.
v = 8.39 × 106 m s−1 (about 2.8% of the speed of light, so ordinary mechanics is still fine here).
Notice the electron is negative and it still gains energy — because it moves towards the higher potential. Negative charges fall “uphill” on a potential diagram.
Potential Due to a Point Charge
This is the building block. Every other potential formula in the chapter is made by adding up copies of this one.
V = (1/4πε₀) × q/r = kq/r
Carry the sign of q into the formula. A positive charge makes the potential around it positive; a negative charge makes it negative.
Where it comes from, in words. The field of a point charge falls off as 1/r2. Potential is the work per unit charge needed to walk in from infinity against that field, and adding up (integrating) a 1/r2 force over distance gives something that goes as 1/r. That single power difference is worth remembering: field goes as 1/r2, potential goes as 1/r. Potential therefore dies away more slowly than field.
Three habits will keep you safe here:
- r is the distance from the charge to the point — always positive, never negative, never a coordinate difference with a sign.
- The sign in the answer comes only from the sign of q.
- V → 0 as r → ∞, which is exactly the zero we chose in the definition.
V = kq/r = (8.99 × 109 N m2 C−2)(4.0 × 10−9 C) / (0.30 m)
Numerator: 8.99 × 4.0 = 35.96, and 109 × 10−9 = 100 = 1, so the numerator is 35.96 N m2 C−1.
Divide by 0.30 m: V = 1.2 × 102 V (119.9 V, sensibly rounded to 120 V).
Unit check: (N m2 C−2)(C)/(m) = N m C−1 = J C−1 = V. Correct.
Step 1. Rearrange: r = kq/V. Both q and V are negative, so the minus signs cancel and r comes out positive — a good sign that you have set it up correctly.
Step 2. r = (8.99 × 109)(2.0 × 10−6) / (9.0 × 104)
Numerator = 1.798 × 104. Divide: 1.798/9.0 = 0.1998, and 104/104 = 1.
Step 3. r = 0.20 m (2 s.f.).
If your r had come out negative, that is the standard warning that you put a signed coordinate in place of a distance.
Step 1 — find V at that point.
V = (8.99 × 109)(6.0 × 10−6)/(0.50) = (5.394 × 104)/(0.50) = 1.0788 × 105 V.
Step 2 — use W = qV, with V(∞) = 0.
W = (1.0 × 10−9 C)(1.0788 × 105 V) = 1.08 × 10−4 J.
Positive work, because you are pushing a positive charge towards another positive charge. Everyday sanity check passed.
Potential Due to a System of Charges
Here is the moment the chapter pays you back for all that vector work last time. To find the field of several charges you had to resolve components and add vectors. To find the potential of several charges you just add numbers.
Plain algebraic addition, signs included. No components. No angles. This is why potential is often the faster route even when the question is really about energy or force.
Two consequences are worth pinning up on your wall. First, potential can be zero at a point where the field is not zero — for instance midway between +q and −q, where the two potentials cancel but the two field vectors point the same way and add. Second, the field can be zero where the potential is not zero — for instance midway between two equal positive charges. Never assume one implies the other.
Step 1. Each charge is r = 0.20 m from the midpoint.
Step 2. V = k[q₁/r + q₂/r] = k(q₁ + q₂)/r, since the distances are equal.
q₁ + q₂ = (+3.0 − 5.0) μC = −2.0 μC = −2.0 × 10−6 C.
Step 3. V = (8.99 × 109)(−2.0 × 10−6)/(0.20) = −1.798 × 104/0.20 = −9.0 × 104 V.
The negative charge is bigger in magnitude, so it wins — the potential is negative. That kind of quick sanity check catches sign errors instantly.
Step 1 — geometry. The centre is half a diagonal from each corner:
r = a/√2 = 0.20/1.4142 = 0.1414 m.
Step 2 — add four equal potentials.
V = 4 × kq/r = 4 × (8.99 × 109)(2.0 × 10−6)/(0.1414)
One term = 1.798 × 104/0.1414 = 1.2714 × 105 V. Times four:
V = 5.1 × 105 V.
Compare this with the field at the centre, which is exactly zero by symmetry — the four vectors cancel in pairs. Same arrangement, huge potential, no field. This is the cleanest possible illustration of the point made above.
Step 1. Let the point be at distance x from the origin, so it is (0.60 − x) from the second charge.
Step 2. Set the total potential to zero:
k(4.0 × 10−6)/x + k(−2.0 × 10−6)/(0.60 − x) = 0
Step 3. Cancel k and 10−6: 4/x = 2/(0.60 − x) ⇒ 4(0.60 − x) = 2x ⇒ 2.40 = 6x.
Step 4. x = 0.40 m from the +4.0 μC charge.
Check: k(4/0.40 − 2/0.20) × 10−6 = k(10 − 10) × 10−6 = 0. It works. (There is a second zero outside the pair too, at x = 1.2 m — try it.)
Potential Due to an Electric Dipole
Recall from the previous chapter that an electric dipole is a pair of equal and opposite charges, +q and −q, separated by a small distance 2a. Its dipole moment is the vector p = q × (2a), pointing from the negative charge to the positive charge, with SI unit C m.
To find the potential at a point P, we do exactly what we did in the last section — add the two point-charge potentials:
V = kq/r₁ + k(−q)/r₂ = kq(1/r₁ − 1/r₂)
where r₁ is the distance from +q to P and r₂ from −q to P. That is exact but clumsy. For a point far away (r much greater than a) the geometry simplifies beautifully and gives the standard result:
where θ is the angle between the dipole moment p and the line drawn from the centre of the dipole to the point P.
On the axis (θ = 0° on the +q side): V = +kp/r2
On the axis (θ = 180°, the −q side): V = −kp/r2
On the equatorial line (θ = 90°): V = 0
Three things to actually understand here, not just memorise.
- The 1/r2 fall-off. A single point charge gives V ∝ 1/r. A dipole is a plus and a minus that almost cancel, so what survives is weaker — it falls as 1/r2. Correspondingly the dipole field falls as 1/r3, one power faster than the potential, exactly as for a point charge.
- Why V = 0 on the equator. Every point on the equatorial plane is equidistant from +q and −q. Equal distances, equal and opposite charges, so the two potentials cancel exactly. This is true even close to the dipole, not just far away.
- V = 0 does not mean E = 0. On the equatorial line the potential vanishes but the field is very much alive — it points antiparallel to p with magnitude kp/r3. Do not let the zero fool you.
(a) Axis, θ = 0°, cosθ = 1.
V = kp/r2 = (8.99 × 109)(4.0 × 10−8)/(0.20)2
Numerator = 3.596 × 102 = 359.6. Denominator = 0.040 m2.
V = 359.6/0.040 = 8.99 × 103 V (about 9.0 kV).
(b) Equator, θ = 90°, cosθ = 0.
V = 0 V, exactly.
Unit check on (a): (N m2 C−2)(C m)/(m2) = N m C−1 = V. Correct.
Step 1. cos 60° = 0.5000.
Step 2. V = kp cosθ/r2 = (8.99 × 109)(6.0 × 10−9)(0.5000)/(0.30)2
Numerator: 8.99 × 6.0 = 53.94, times 109 × 10−9 = 1, times 0.5 = 26.97.
Denominator: (0.30)2 = 0.090 m2.
Step 3. V = 26.97/0.090 = 3.0 × 102 V (299.7 V).
Because cos60° = ½, this is exactly half the axial value at the same distance. Handy mental check.
Equipotential Surfaces
Think of a contour map of a hill. Every line on it joins points of the same height, and walking along a contour costs you no effort against gravity. An equipotential surface is exactly that idea for electricity: a surface on which the potential has the same value everywhere.
2. The electric field is always perpendicular to the equipotential surface, pointing from higher to lower potential.
3. Two equipotential surfaces can never intersect. If they did, the point of intersection would have two different potentials at once, which is impossible.
4. Equipotentials are closer together where the field is stronger, since a strong field means the potential changes rapidly with distance.
Why must E be perpendicular? This is the one that gets asked. Suppose the field had a component along the surface. Then moving a charge a little way along the surface would involve a force component in the direction of motion, so work would be done — but moving along an equipotential means ΔV = 0, so the work must be zero. The only escape is that the along-the-surface component of E is zero. Hence E is entirely perpendicular. Write that argument out once and you own a two-mark answer forever.
What they look like. For an isolated point charge they are concentric spheres centred on the charge. For a uniform field (as between the plates of a capacitor) they are flat planes perpendicular to the field. For a dipole they are two families of closed surfaces wrapped around the two charges, with the equatorial plane itself being the V = 0 surface. And the surface of any charged conductor in equilibrium is an equipotential — more on that shortly.
Step 1. V = kq/r, so r = kq/V. Compute kq once:
kq = (8.99 × 109)(2.0 × 10−9) = 17.98 V m.
Step 2. Divide by each potential:
r(100 V) = 17.98/100 = 0.180 m
r(50 V) = 17.98/50 = 0.360 m
r(25 V) = 17.98/25 = 0.719 m
Comment. Halving the potential doubles the radius. So going from the 100 V surface to the 50 V surface takes 0.180 m, but from 50 V to 25 V takes 0.359 m — the surfaces get further apart as you move out. Property 4 in action: the field is weaker further away, so you must travel further for the same drop in potential.
Model answer. Zero. Work done is W = q(Vfinal − Vinitial). Every point of the path lies on the same equipotential surface, so Vfinal = Vinitial and ΔV = 0. Therefore W = q × 0 = 0 J, whatever the size or shape of the path. (In fact the work around any closed path in an electrostatic field is zero, because the electrostatic force is conservative.)
Notice the structure: formula → reason → numerical conclusion → one extra sentence of physics. That is how full marks look.
Relation Between Field and Potential
Field and potential are two descriptions of the same physical situation, so there has to be a bridge between them. Here it is.
In one dimension: Ex = −dV/dx. For a uniform field over a distance d: E = V/d.
Because of this, the volt per metre (V m−1) is an alternative SI unit for electric field, completely equivalent to the newton per coulomb (N C−1).
Where the minus sign comes from. Move a unit positive charge a small distance dr along the field. The field does work E·dr on it, so the charge loses potential energy, meaning the potential drops: dV = −E dr. Rearranged, E = −dV/dr. Physically the minus sign says: the electric field points downhill on the potential landscape, from high V to low V. A positive charge released from rest always rolls towards lower potential, just as a ball rolls downhill.
The quantity dV/dr is called the potential gradient. Steep gradient means strong field; flat gradient means weak field; zero gradient (constant V) means zero field. That last one is exactly why the field inside a conductor is zero — the potential inside is constant.
Step 1. Differentiate: dV/dx = 10x (the constant 3 differentiates to zero — a constant added to a potential never affects the field, which is the mathematical version of “only differences matter”).
Step 2. Ex = −dV/dx = −10x.
Step 3. At x = 2.0 m: Ex = −10(2.0) = −20 V m−1.
Reading the answer: the magnitude is 20 V m−1 and the minus sign means it points in the −x direction. That makes sense — V increases with x here, so the downhill direction is backwards along x.
Step 1 — the field. d = 2.0 mm = 2.0 × 10−3 m.
E = V/d = 60/(2.0 × 10−3) = 3.0 × 104 V m−1.
Step 2 — the force. F = qE = (1.602 × 10−19 C)(3.0 × 104 N C−1) = 4.8 × 10−15 N, directed from the low-potential plate towards the high-potential plate (electrons are attracted to the positive plate).
This E = V/d shortcut only works because the field between closely spaced parallel plates is uniform. Never use it for a point charge.
Potential Energy of a System of Two Point Charges
So far we have moved a test charge around in someone else’s field. Now we ask a different question: how much energy is locked up in the arrangement itself?
Imagine building the system from scratch, with all the charges starting infinitely far apart. Bringing in the first charge costs nothing — there is nothing to push against. Bringing in the second charge costs q₂ multiplied by the potential that charge q₁ has already set up at that spot. That total cost is the electrostatic potential energy of the system.
Three charges: add one term for every pair —
U = k[ q₁q₂/r₁₂ + q₂q₃/r₂₃ + q₁q₃/r₁₃ ]
n charges: there are n(n−1)/2 pairs. For 3 charges that is 3 terms, for 4 charges it is 6 terms.
Signs go straight in. Unit: joule. U is a scalar, and U → 0 when all separations → ∞.
Reading the sign of U. If U is positive (like charges), the system is storing energy and would fly apart if released — you had to do work to assemble it. If U is negative (unlike charges), the system is bound; the charges pulled themselves together and you would have to supply that much energy to separate them to infinity. This is exactly the logic used later for binding energy in atoms and nuclei, so it is worth getting fluent now.
U = kq₁q₂/r = (8.99 × 109)(3.0 × 10−6)(4.0 × 10−6)/(0.20)
Charges: (3.0)(4.0) = 12, and 10−6 × 10−6 = 10−12, so q₁q₂ = 1.2 × 10−11 C2.
Numerator: (8.99 × 109)(1.2 × 10−11) = 0.10788 J m.
U = 0.10788/0.20 = +0.54 J.
Positive, as expected for two positive charges. It also means: release them and they will fly apart, sharing 0.54 J of kinetic energy between them.
Step 1 — count the pairs. n = 3, so n(n−1)/2 = 3 pairs. By symmetry all three pairs are identical.
Step 2. U = 3 × kq2/a = 3 × (8.99 × 109)(1.0 × 10−6)2/(0.10)
q2 = 1.0 × 10−12 C2. One term = (8.99 × 109)(1.0 × 10−12)/0.10 = 8.99 × 10−2 J.
Step 3. U = 3 × 0.0899 = +0.27 J.
Watch out: pairs, not charges. A very common error is to write U = 3 × something and then also divide by 2 “to avoid double counting”. Once you count each pair exactly once, no extra factor is needed.
Step 1 — initial energy. q₁q₂ = (2.0 × 10−6)(−3.0 × 10−6) = −6.0 × 10−12 C2.
Ui = (8.99 × 109)(−6.0 × 10−12)/(0.30) = (−5.394 × 10−2)/0.30 = −0.1798 J.
Step 2 — final energy. Doubling r halves the magnitude:
Uf = −0.0899 J.
Step 3 — work done. Wext = Uf − Ui = (−0.0899) − (−0.1798) = +0.090 J.
Positive work, because the charges attract and you are pulling them apart. Notice that both energies are negative yet the change is positive — keeping careful track of signs in the subtraction is the whole trick here.
Potential Energy of a Dipole in a Uniform Field
Put a dipole in a uniform external field E. The two charges feel equal and opposite forces, so the net force is zero — the dipole does not drift. But those forces act at different points, so there is a torque that tries to twist the dipole into line with the field.
Potential energy: U = −pE cosθ = −p·E, unit J
Work to rotate from θ₁ to θ₂: W = U₂ − U₁ = pE(cosθ₁ − cosθ₂)
Here θ is the angle between p and E. The zero of this potential energy is taken at θ = 90°.
Make sense of the three special angles and you will never need to memorise anything else here:
| Angle θ between p and E | Torque τ | Energy U | State |
|---|---|---|---|
| 0° (p along E) | 0 | −pE (minimum) | Stable equilibrium |
| 90° (perpendicular) | pE (maximum) | 0 | Maximum twisting effect |
| 180° (p opposite E) | 0 | +pE (maximum) | Unstable equilibrium |
Compare it to a compass needle in the earth’s magnetic field. Lined up with the field it sits happily — lowest energy, no torque. Turned sideways it twists hardest. Turned right around it balances for a moment but the tiniest nudge flips it. Same physics, same picture.
Step 1 — compute pE once.
pE = (5.0 × 10−8 C m)(2.0 × 104 N C−1) = 1.0 × 10−3 N m = 1.0 × 10−3 J.
Step 2 — torque. sin60° = 0.8660.
τ = pE sinθ = (1.0 × 10−3)(0.8660) = 8.7 × 10−4 N m.
Step 3 — energy. cos60° = 0.5000.
U = −pE cosθ = −(1.0 × 10−3)(0.5000) = −5.0 × 10−4 J.
Negative energy tells you the dipole is on the “attracted” side — it is closer to lining up than to being flipped, and the torque is trying to close that last 60°.
Use W = pE(cosθ₁ − cosθ₂) with pE = 1.0 × 10−3 J.
(a) 0° → 90°: W = pE(cos0° − cos90°) = pE(1 − 0) = 1.0 × 10−3 J.
(b) 0° → 180°: W = pE(cos0° − cos180°) = pE(1 − (−1)) = 2pE = 2.0 × 10−3 J.
So flipping a dipole completely over costs exactly twice as much as turning it sideways. Both answers are positive because you are always rotating away from the stable position.
Conductors in an Electrostatic Field: Free and Bound Charges
This section is almost entirely theory, and almost entirely free marks. The whole thing follows from one sentence: in a conductor there are charges that can move, and they will keep moving until they have no reason to move any more.
Free charges are the conduction electrons in a metal — roughly one per atom, detached from their parent atoms and free to wander through the whole lump of metal. Bound charges are the electrons still tightly held to their atoms plus the positive ions of the lattice; they can shift by a tiny fraction of an atomic diameter but cannot travel. In an insulator, all the charges are bound — that is exactly what makes it an insulator.
2. Any excess charge sits entirely on the outer surface. Apply Gauss’s theorem to a surface drawn just inside the conductor: E = 0 on it, so the enclosed charge must be zero. So there is no charge in the bulk.
3. The whole conductor is at one potential. Since E = −dV/dr and E = 0 inside, V cannot change — the interior and the surface are all at the same potential. The conductor is an equipotential volume.
4. Just outside the surface, E is perpendicular to the surface with magnitude E = σ/ε₀, where σ is the local surface charge density. Any parallel component would push surface charges sideways.
5. Charge density is largest where the surface is sharpest. On a pointed conductor σ is huge at the tip, so the field there is huge — this is the principle behind the lightning conductor and behind corona discharge.
(a) and (b) — inside. E = 0 at both points, by property 1. The potential is the same as at the surface, by property 3.
(c) — at the surface, the sphere behaves like a point charge at its centre:
V = kq/R = (8.99 × 109)(5.0 × 10−9)/(0.10) = (44.95)/(0.10) = 4.5 × 102 V (449.5 V).
E at the surface = kq/R2 = 44.95/0.010 = 4.5 × 103 V m−1.
So the answers to (a) and (b) are: E = 0, V = 4.5 × 102 V at both points. A perfect illustration that zero field does not mean zero potential.
Step 1 — the physical principle. Connected conductors form one conductor, so they must end at the same potential. Charge flows until that is true.
Step 2 — set the potentials equal. kq₁/R₁ = kq₂/R₂ ⇒ q₁/q₂ = R₁/R₂ = 0.050/0.10 = 1/2.
Step 3 — split 9.0 nC in the ratio 1 : 2.
q₁ = 3.0 nC on the small sphere, q₂ = 6.0 nC on the large one.
Step 4 — common potential. V = (8.99 × 109)(3.0 × 10−9)/(0.050) = 5.4 × 102 V (539.4 V).
Cross-check on the big sphere: (8.99 × 109)(6.0 × 10−9)/(0.10) = 539.4 V. Identical, as required.
Step 5 — surface charge densities. σ = q/4πR2, so σ₁/σ₂ = (q₁/q₂)(R₂/R₁)2 = (1/2)(2)2 = 2 : 1.
The lesson: the smaller sphere ends up with less charge but a higher charge density — in general σ ∝ 1/R. That is property 5 above, and it is the reason lightning conductors are pointed.
Dielectrics and Electric Polarisation
A dielectric is just an insulator viewed through the lens of what it does in an electric field. Glass, mica, paper, oil, polythene, even dry air. It has no free charges, so no current flows through it — but its bound charges can still shift very slightly, and that tiny shift has big consequences.
Dielectrics come in two flavours:
- Non-polar (H₂, O₂, CO₂, most plastics): the centres of positive and negative charge coincide, so each molecule has no permanent dipole moment. Switch on a field and the electron cloud gets tugged one way, the nucleus the other, and an induced dipole moment appears in the direction of the field.
- Polar (H₂O, HCl, NH₃): the molecule is permanently lopsided and already has a dipole moment. Without a field, thermal jostling points them randomly in all directions, so the bulk moment is zero. Switch on a field and they partly align, and again a net moment appears along the field.
Either way, the end result is the same: the slab develops a net dipole moment along the field. That is electric polarisation.
For a linear isotropic dielectric: P = ε₀ χe E, where χe is the electric susceptibility.
The aligned dipoles leave uncancelled bound surface charges: negative on the face where the field enters, positive on the face where it leaves. Their field opposes the applied field, so inside the dielectric
E = E₀/K where K (also written εr) is the dielectric constant.
K = 1 for vacuum, about 1.0006 for air, roughly 2 for polythene, 6 for mica, 80 for water. K is a pure number with no units, and K ≥ 1 always. Also χe = K − 1, and the bound surface charge density equals P.
The mental picture that makes this stick. Imagine a crowd of tiny compass needles frozen in jelly. Switch on a field and they all swing to point the same way. In the middle of the slab every needle’s head sits next to the next needle’s tail, so the charges cancel. Only at the two ends of the slab is there nothing to cancel against — and that is precisely where the bound surface charge appears. Those two sheets of bound charge produce a field pointing backwards, weakening the original field by the factor K.
Note the crucial difference from a conductor: a conductor reduces the internal field to exactly zero, because its free charges keep moving until they have completely cancelled the field. A dielectric only reduces it by the factor K, because the bound charges can shift a little but cannot travel.
(a) E = E₀/K = (1.2 × 104)/4.0 = 3.0 × 103 V m−1.
(b) P = ε₀χeE = ε₀(K − 1)E
= (8.854 × 10−12 F m−1)(4.0 − 1)(3.0 × 103 V m−1)
= (8.854 × 10−12)(3.0)(3.0 × 103) = 7.97 × 10−8 C m−2.
(c) The magnitude of the bound surface charge density equals P, so σb = 7.97 × 10−8 C m−2 — negative on the entry face, positive on the exit face.
Unit check on (b): (F m−1)(V m−1) = (C V−1 m−1)(V m−1) = C m−2. Exactly a surface charge density, as it should be.
Model answer. In a non-polar dielectric the centres of positive and negative charge of each molecule coincide, so the molecules have zero permanent dipole moment (examples: O₂, CO₂). In a polar dielectric the two centres do not coincide, so each molecule has a permanent dipole moment (examples: H₂O, HCl); however, thermal agitation orients them randomly, so the net moment of the bulk material is still zero.
When an external field E₀ is applied, non-polar molecules are stretched and acquire an induced dipole moment along E₀, while polar molecules experience a torque and partially align with E₀. In both cases the dielectric acquires a net dipole moment per unit volume, called polarisation P, in the direction of E₀. Uncancelled bound charges appear on the two surfaces, and their field opposes E₀, reducing the field inside to E = E₀/K.
Capacitors and Capacitance
Here is the second big idea of the chapter. Give an isolated conductor some charge and its potential rises. Give it more charge and the potential rises more — and experiment shows the two are strictly proportional: Q ∝ V. The constant of proportionality is called the capacitance.
SI unit: the farad (F), where 1 F = 1 C V−1.
For a capacitor (two conductors carrying +Q and −Q), Q is the magnitude of the charge on either plate and V is the potential difference between them.
C depends only on geometry and on the medium — on the size, shape and spacing of the conductors and on the dielectric between them. It does not depend on how much charge you happen to have put on it. Double Q and V doubles too, leaving C unchanged.
An analogy that actually helps. Think of a bucket. Q is the volume of water you pour in, V is the height the water reaches, and C is the cross-sectional area of the bucket. A wide bucket (large C) takes a lot of water for very little rise in level. A narrow test tube (small C) shoots up in level with barely any water. And the “area of the bucket” is fixed by its shape, not by how much water is in it — exactly like capacitance.
The farad is enormous. One farad means storing one coulomb at one volt, and a coulomb is a colossal amount of static charge. Real capacitors are therefore labelled in submultiples:
| Name | Symbol | In farads | Typical use |
|---|---|---|---|
| microfarad | μF | 10−6 F | Power supplies, fan motors |
| nanofarad | nF | 10−9 F | Timing and filter circuits |
| picofarad | pF | 10−12 F | Radio tuning, oscillators |
(a) C = Q/V = (6.0 × 10−6 C)/(12 V) = 5.0 × 10−7 F = 0.50 μF.
(b) Q = CV = (20 × 10−6 F)(200 V) = 4.0 × 10−3 C = 4.0 mC.
Shortcut worth knowing: microfarads × volts = microcoulombs. In (b), 20 μF × 200 V = 4000 μC = 4.0 mC. Same answer, no powers of ten to lose.
The Parallel Plate Capacitor
Two flat metal plates, each of area A, held a small distance d apart, facing each other. Put +Q on one and −Q on the other. That is the capacitor you will be asked about again and again, so let us derive its capacitance properly — it is a three-line derivation and it is worth marks.
Step 1 — the field between the plates. The surface charge density on each plate is σ = Q/A. From Gauss’s theorem, the field between two oppositely charged parallel sheets is uniform and equal to
E = σ/ε₀ = Q/(ε₀A)
Step 2 — the potential difference. The field is uniform, so V = Ed:
V = Qd/(ε₀A)
Step 3 — divide. C = Q/V = Q ÷ [Qd/(ε₀A)], and the Q cancels — which is exactly why C does not depend on charge:
Bigger plates → bigger C (more room to spread charge out, so V rises less).
Smaller gap → bigger C (the opposite plate pulls harder, so more charge can sit there for the same V).
This formula assumes the plates are large compared with d, so the field is uniform and we can ignore the “fringing” that bulges out at the edges.
Step 1 — convert everything to SI. This is where marks are lost.
A = 100 cm2 = 100 × (10−2 m)2 = 100 × 10−4 m2 = 1.0 × 10−2 m2
d = 1.0 mm = 1.0 × 10−3 m
Step 2 — substitute.
C = ε₀A/d = (8.854 × 10−12)(1.0 × 10−2)/(1.0 × 10−3)
= 8.854 × 10−14/10−3 = 8.854 × 10−11 F
Step 3 — express it sensibly. C = 88.5 pF.
Note the conversion trap: 1 cm2 = 10−4 m2, not 10−2 m2. Squaring the centimetre squares the conversion factor too.
d = ε₀A/C = (8.854 × 10−12 F m−1)(1.0 m2)/(1.0 F) = 8.9 × 10−12 m.
That is about 8.9 picometres — roughly a tenth of the diameter of a single atom. Physically impossible. This is the cleanest way to feel just how big the farad is, and it is a lovely one-liner if an examiner asks you to comment on the size of the SI unit.
(a) C = ε₀A/d = (8.854 × 10−12)(0.020)/(2.0 × 10−3) = (1.7708 × 10−13)/(2.0 × 10−3) = 8.85 × 10−11 F = 88.5 pF.
(b) Q = CV = (8.854 × 10−11)(100) = 8.85 × 10−9 C = 8.85 nC.
(c) E = V/d = 100/(2.0 × 10−3) = 5.0 × 104 V m−1.
(d) σ = Q/A = (8.854 × 10−9)/(0.020) = 4.43 × 10−7 C m−2.
Cross-check: σ/ε₀ should equal E. (4.427 × 10−7)/(8.854 × 10−12) = 5.0 × 104 V m−1. It matches part (c) exactly. Whenever a question gives you several parts like this, use one part to check another — it costs ten seconds and catches almost every arithmetic slip.
Effect of a Dielectric on Capacitance
Now put the two halves of the chapter together. Slide a dielectric slab into the gap of a parallel plate capacitor. The bound charges polarise, the field inside the slab drops by the factor K, so the potential difference across the plates drops — and since C = Q/V, a smaller V for the same Q means a larger C.
C = Kε₀A/d = K C₀
Slab of thickness t inserted into a gap of width d (t < d):
C = ε₀A / (d − t + t/K)
Conducting slab of thickness t inserted (equivalent to K → ∞):
C = ε₀A / (d − t)
Notice how the general formula contains the other two. Put t = d and you get Kε₀A/d. Let K → ∞ so that t/K → 0 and you get ε₀A/(d − t). Learn the middle one and the other two come free.
Now the part that decides whole questions: was the battery left connected or not? Everything hinges on which quantity is held fixed, so before you write anything, find that sentence in the question and underline it.
| Quantity | Battery still connected (V fixed) | Battery disconnected (Q fixed) |
|---|---|---|
| Capacitance C | becomes K times larger | becomes K times larger |
| Potential difference V | unchanged (the battery holds it) | falls to V/K |
| Charge Q | rises to KQ (more flows in) | unchanged (nowhere to go) |
| Field E between plates | unchanged (E = V/d, both fixed) | falls to E/K |
| Energy U | rises to KU (use U = ½CV2) | falls to U/K (use U = Q2/2C) |
The trick for remembering which energy formula to use. Pick the version of the energy formula written in terms of the quantity that is held constant. Battery connected means V is constant, so use U = ½CV2 and only C changes. Battery disconnected means Q is constant, so use U = Q2/2C and again only C changes. Choose the right form and the answer takes one line instead of five.
Before insertion: C₀ = 8.854 × 10−11 F, V₀ = 100 V,
Q₀ = C₀V₀ = 8.854 × 10−9 C, U₀ = ½C₀V₀2 = 4.43 × 10−7 J.
In both cases: C = KC₀ = 5.0 × 8.854 × 10−11 = 4.43 × 10−10 F (442.7 pF).
(a) Battery connected — V stays at 100 V.
Q = CV = (4.427 × 10−10)(100) = 4.43 × 10−8 C (five times as much charge; the extra came from the battery).
U = ½CV2 = ½(4.427 × 10−10)(100)2 = 2.21 × 10−6 J = 5U₀. The battery supplied that extra energy.
E = V/d = 100/(1.0 × 10−3) = 1.0 × 105 V m−1, unchanged.
(b) Battery disconnected — Q stays at 8.854 × 10−9 C.
V = Q/C = (8.854 × 10−9)/(4.427 × 10−10) = 20 V (= 100/5, as expected).
U = Q2/2C = (8.854 × 10−9)2/(2 × 4.427 × 10−10) = 8.85 × 10−8 J = U₀/5.
E = V/d = 20/(1.0 × 10−3) = 2.0 × 104 V m−1 = E₀/5.
Where did the lost energy go in (b)? The capacitor pulled the slab in — the electric field does work on the slab, so the stored energy falls. If you let go of the slab near the edge of the gap, it is sucked in. That is a genuine, observable effect, not a bookkeeping trick.
Step 1 — before.
C₀ = ε₀A/d = (8.854 × 10−12)(5.0 × 10−3)/(4.0 × 10−3) = (4.427 × 10−14)/(4.0 × 10−3) = 1.11 × 10−11 F (11.1 pF).
Step 2 — the effective gap after insertion.
d − t + t/K = 4.0 × 10−3 − 2.0 × 10−3 + (2.0 × 10−3)/4.0
= 2.0 × 10−3 + 0.50 × 10−3 = 2.5 × 10−3 m.
Step 3 — after.
C = (4.427 × 10−14)/(2.5 × 10−3) = 1.77 × 10−11 F (17.7 pF).
Ratio C/C₀ = 17.7/11.1 = 1.6. Only a 60% increase, even though K = 4. That is because the slab fills just half the gap — the air half still dominates. A nice reality check on the formula.
Bonus: the position of the slab in the gap makes no difference at all. Only its thickness matters. Try it in the formula and see.
Capacitors in Series and in Parallel
Real circuits contain several capacitors, and we usually want the single “equivalent” capacitor that would behave identically. There are two basic ways to wire them, and each has one governing rule you must never mix up.
Series: end to end, one path. Why must the charge be the same on all of them? Look at the isolated conductor formed by the right-hand plate of C₁, the wire, and the left-hand plate of C₂. It was neutral before, and no charge can get on or off it, so whatever −Q appears on one plate must be matched by +Q on the other. That argument repeats down the chain, forcing the same Q everywhere. Meanwhile the potential drops add up as you walk from one battery terminal to the other: V = V₁ + V₂ + V₃.
Substituting Vi = Q/Ci and V = Q/C into that sum, the common Q cancels and leaves the reciprocal rule.
Parallel: side by side, all across the same two nodes. Both plates of every capacitor are joined to the same pair of terminals, so every capacitor has exactly the same potential difference V. The total charge drawn from the battery is shared out: Q = Q₁ + Q₂ + Q₃. Substituting Qi = CiV and Q = CV, the common V cancels and you are left with a simple sum.
Parallel: C = C₁ + C₂ + C₃ + … — the equivalent C is larger than the largest individual capacitor.
Two in series is worth memorising directly: C = C₁C₂/(C₁ + C₂), the “product over sum” form.
(a) Series. 1/C = 1/2 + 1/3 + 1/6. Use a common denominator of 6:
1/C = 3/6 + 2/6 + 1/6 = 6/6 = 1 (μF)−1
C = 1.0 μF. Smaller than the smallest (2.0 μF). Check passed.
(b) Parallel. C = 2.0 + 3.0 + 6.0 = 11 μF. Larger than the largest (6.0 μF). Check passed.
Note the factor of eleven between the two answers — how you wire capacitors matters enormously. Also note you may work entirely in μF as long as every capacitor is in μF; the answer then comes out in μF.
Step 1 — the charge. In series every capacitor carries the same Q, and that Q is set by the equivalent capacitance:
Q = CeqV = (1.0 × 10−6 F)(12 V) = 1.2 × 10−5 C = 12 μC on every capacitor.
Step 2 — the individual voltages, using Vi = Q/Ci:
V₁ = 12 μC / 2.0 μF = 6.0 V
V₂ = 12 μC / 3.0 μF = 4.0 V
V₃ = 12 μC / 6.0 μF = 2.0 V
Step 3 — check. 6.0 + 4.0 + 2.0 = 12 V. Matches the battery exactly. Always do this check; it catches errors instantly.
Read the pattern: the smallest capacitor takes the biggest share of the voltage. V ∝ 1/C in a series chain. This is why a small capacitor in a series string is the one most at risk of breaking down.
Step 1 — always start with the innermost group. The parallel pair:
CP = 2.0 + 6.0 = 8.0 μF.
Step 2 — now that block is in series with 4.0 μF. Use product over sum:
C = (4.0 × 8.0)/(4.0 + 8.0) = 32/12 = 2.7 μF (2.667 μF exactly).
Sanity check: the final answer is smaller than 4.0 μF, which it must be, because the last step was a series step. If you had accidentally added, you would have got 12 μF and the check would have caught you.
Method note: redraw the circuit after every step. One simplified sketch per step turns a scary five-capacitor network into three easy ones.
Energy Stored in a Capacitor
Charging a capacitor is not free. To move charge onto a plate that already carries charge of the same sign, you have to push against the repulsion — and that work does not disappear. It is stored in the capacitor, ready to be released the moment you complete a circuit. That is why a camera flash can dump a huge burst of energy in a millisecond even though a small battery charged it slowly.
All three are the same energy written differently, linked by Q = CV. Unit: joule.
The 2026-27 Class 12 Physics curriculum lists this topic as “energy stored in a capacitor — no derivation, formulae only”. So you will not be asked to derive it. You will be asked to use it, often in questions where something about the capacitor changes. That is exactly what we practise below.
How to choose between the three forms — the only skill you need here. Look at what the question tells you and, crucially, at what stays constant:
| You are given / holding constant | Use this form | Because |
|---|---|---|
| C and V (battery connected) | U = ½CV2 | V is fixed, so U tracks C directly |
| Q and C (battery disconnected) | U = Q2/2C | Q is fixed, so U falls as C rises |
| Q and V (and C not asked for) | U = ½QV | No need to find C at all |
U = ½CV2 = ½(50 × 10−6 F)(200 V)2
(200)2 = 4.0 × 104 V2
U = 0.5 × (50 × 10−6) × (4.0 × 104) = 0.5 × 2.0 = 1.0 J
Unit check: F × V2 = (C V−1)(V2) = C V = J. Correct. And note the V2 — doubling the voltage quadruples the stored energy. Squaring is easy to forget in a hurry.
Route 1 — U = Q2/2C.
Q2 = (3.0 × 10−4)2 = 9.0 × 10−8 C2
2C = 2(25 × 10−6) = 5.0 × 10−5 F
U = (9.0 × 10−8)/(5.0 × 10−5) = 1.8 × 10−3 J = 1.8 mJ
Route 2 — U = ½QV. First V = Q/C = (3.0 × 10−4)/(25 × 10−6) = 12 V.
U = ½(3.0 × 10−4)(12) = 1.8 × 10−3 J. Identical.
Two independent routes agreeing to the last digit is the strongest confidence you can get in an exam.
Step 1 — charge is conserved.
Q = C₁V₁ = (2.0 × 10−6)(100) = 2.0 × 10−4 C. Nothing can escape, so this total is fixed.
Step 2 — they end up in parallel, so they share one common V.
V = Q/(C₁ + C₂) = (2.0 × 10−4)/(5.0 × 10−6) = 40 V.
Step 3 — energies.
Ui = ½C₁V₁2 = ½(2.0 × 10−6)(100)2 = 1.0 × 10−2 J
Uf = ½(C₁ + C₂)V2 = ½(5.0 × 10−6)(40)2 = 4.0 × 10−3 J
Energy lost = 1.0 × 10−2 − 4.0 × 10−3 = 6.0 × 10−3 J (6.0 mJ, which is 60% of the original).
Where did it go? Not lost from the universe — it is dissipated as heat in the connecting wires (and a little as electromagnetic radiation) while the charge surges across. Energy is always lost in this kind of sharing, however good the wires are. If you want the standard shortcut: energy lost = C₁C₂(V₁ − V₂)2 / [2(C₁ + C₂)] = (2 × 3 × 10−12)(100)2/(2 × 5 × 10−6) = 6.0 × 10−3 J. Same answer.
u = ½ε₀E2 (joules per cubic metre)
Quick example 37: for E = 1.0 × 105 V m−1,
u = ½(8.854 × 10−12)(1.0 × 105)2 = ½(8.854 × 10−12)(1.0 × 1010) = 4.4 × 10−2 J m−3.
Multiply u by the volume Ad of the gap and you recover ½CV2 exactly — a satisfying consistency check, and a nice extra line if a question asks you to comment.
Formula Sheet for Revision
Copy this table onto one side of a sheet of paper by hand. Do not photograph it — the act of writing it out is what makes it stick. Read it the night before the exam and again on the morning.
| Quantity | Formula | SI unit |
|---|---|---|
| Work and potential difference | W = qΔV | J |
| Potential of a point charge | V = kq/r | V |
| Potential of a system | V = kΣqi/ri | V |
| Potential of a short dipole | V = kp cosθ/r2 | V |
| Charged conducting sphere | V = kq/r (r ≥ R); V = kq/R (r ≤ R) | V |
| Field from potential | E = −dV/dr; uniform field E = V/d | V m−1 |
| PE of two point charges | U = kq₁q₂/r | J |
| PE of n charges | one term per pair; n(n−1)/2 pairs | J |
| Dipole in a uniform field | τ = pE sinθ; U = −pE cosθ | N m; J |
| Work to rotate a dipole | W = pE(cosθ₁ − cosθ₂) | J |
| Field just outside a conductor | E = σ/ε₀, perpendicular to the surface | V m−1 |
| Polarisation | P = ε₀(K − 1)E; E = E₀/K | C m−2 |
| Capacitance | C = Q/V | F |
| Parallel plate capacitor | C = ε₀A/d; with full dielectric C = Kε₀A/d | F |
| Partly filled with a slab | C = ε₀A/(d − t + t/K) | F |
| Series combination | 1/C = Σ1/Ci; same Q, voltages add | F |
| Parallel combination | C = ΣCi; same V, charges add | F |
| Energy stored (formulae only) | U = ½CV2 = ½QV = Q2/2C | J |
| Energy density of the field | u = ½ε₀E2 | J m−3 |
Practice Worksheet
Ten questions, ranging from one-liners to full board-style numericals. Do them with the answers covered, on paper, with a calculator and your formula sheet. Only then open the reveals. If you get one wrong, do not just read the solution — close it and redo the question from scratch.
Q1. Two point charges, +5.0 nC and −3.0 nC, are fixed 0.20 m apart. Find the electric potential at the midpoint of the line joining them.
Show Answer
qtotal = (+5.0 − 3.0) nC = +2.0 nC = 2.0 × 10−9 C.
V = kq/r = (8.99 × 109)(2.0 × 10−9)/(0.10) = 17.98/0.10 = +1.8 × 102 V (179.8 V).
Note that the field at this point is certainly not zero — both field vectors point from the positive towards the negative charge, so they add.
Q2. An alpha particle (charge +2e) starts from rest and is accelerated through a potential difference of 500 V. Find its kinetic energy in electron volts and in joules.
Show Answer
In joules: KE = (2 × 1.602 × 10−19 C)(500 V) = (3.204 × 10−19)(500) = 1.60 × 10−16 J.
Check: 1000 eV × 1.602 × 10−19 J/eV = 1.602 × 10−16 J. Agrees.
Q3. A short electric dipole has dipole moment 2.0 × 10−9 C m. Find the potential at a point 0.10 m from its centre (a) on the axis, on the positive side, and (b) on the equatorial line. Explain the difference physically.
Show Answer
V = kp/r2 = (8.99 × 109)(2.0 × 10−9)/(0.10)2 = 17.98/0.010 = 1.8 × 103 V (1798 V).
(b) θ = 90°, cosθ = 0, so V = 0 V.
Why: every point on the equatorial plane is the same distance from +q and from −q, so the two potentials are equal in magnitude and opposite in sign and cancel exactly. On the axis you are nearer one charge than the other, so no cancellation occurs.
Q4. A parallel plate capacitor has plates of area 250 cm2 separated by 2.0 mm of air. Find its capacitance, and its new capacitance if the gap is completely filled with a dielectric of K = 3.0.
Show Answer
C₀ = ε₀A/d = (8.854 × 10−12)(2.5 × 10−2)/(2.0 × 10−3)
= (2.2135 × 10−13)/(2.0 × 10−3) = 1.11 × 10−10 F = 110.7 pF.
With the dielectric: C = KC₀ = 3.0 × 1.107 × 10−10 = 3.32 × 10−10 F = 332 pF.
Q5. Three capacitors of 4.0 μF, 4.0 μF and 2.0 μF are available. Find the equivalent capacitance when they are connected (a) all in series, (b) all in parallel, and state which arrangement stores more energy at the same supply voltage.
Show Answer
(b) Parallel. C = 4.0 + 4.0 + 2.0 = 10 μF.
(c) At the same supply voltage, U = ½CV2, so the larger C stores more. The parallel arrangement stores ten times as much energy as the series one.
Q6. A 100 μF capacitor is charged to 50 V. Find the charge stored and the energy stored. If the voltage is then doubled to 100 V, by what factor does each change?
Show Answer
U = ½CV2 = ½(100 × 10−6)(50)2 = ½(100 × 10−6)(2500) = 0.125 J.
Doubling V: Q doubles (Q ∝ V, so 10 mC) but U becomes four times larger (U ∝ V2, so 0.500 J). The capacitance itself does not change at all — it is fixed by geometry and the dielectric.
Q7. Point A is at a potential of −30 V and point B at +50 V. How much work must an external agent do to carry a charge of 2.0 μC slowly from A to B? Does the answer depend on the path taken?
Show Answer
W = qΔV = (2.0 × 10−6 C)(80 V) = 1.6 × 10−4 J.
No, the path makes no difference. The electrostatic force is conservative, so the work done depends only on the endpoints. The most common slip here is writing 50 − 30 = 20 V; take real care with the double minus.
Q8. Two charges of +2.0 μC each are held 0.50 m apart. (a) Find the potential energy of the system. (b) What does the sign tell you? (c) How much energy must be supplied to move them infinitely far apart?
Show Answer
q₁q₂ = 4.0 × 10−12 C2; numerator = 3.596 × 10−2 J m; U = 0.03596/0.50 = +7.2 × 10−2 J (0.0719 J).
(b) Positive U means work had to be done against repulsion to assemble the system. Released, the charges would fly apart and share 0.072 J of kinetic energy.
(c) U at infinity is zero, so the energy needed is U∞ − U = 0 − 0.0719 = −0.072 J — that is, no energy needs to be supplied at all. The system does 0.072 J of work on you as it flies apart. (Contrast this with two unlike charges, where you would genuinely have to supply energy.)
Q9. A 10 μF capacitor is charged to 200 V and then disconnected from the battery. A dielectric of K = 2.0 is then inserted, completely filling the gap. Find the new capacitance, the new potential difference, and the energy before and after. Account for any change in energy.
Show Answer
Q = C₀V₀ = (10 × 10−6)(200) = 2.0 × 10−3 C.
New capacitance: C = KC₀ = 2.0 × 10 μF = 20 μF.
New potential difference: V = Q/C = (2.0 × 10−3)/(20 × 10−6) = 100 V (= V₀/K, as expected).
Ui = ½C₀V₀2 = ½(10 × 10−6)(200)2 = 0.20 J
Uf = ½CV2 = ½(20 × 10−6)(100)2 = 0.10 J
Accounting: the energy halves (Uf = Ui/K). The “missing” 0.10 J is the work done by the electric field in pulling the slab into the gap. If you held the slab back, you would feel it being tugged in, and you would have to absorb that 0.10 J.
Q10. A uniform field of 500 V m−1 points along the +x direction. (a) Find the potential difference V(x = 0.20 m) − V(x = 0.60 m). (b) State two properties of equipotential surfaces and describe their shape in this field.
Show Answer
V(0.20) − V(0.60) = EΔx = (500 V m−1)(0.40 m) = +200 V.
The point at x = 0.20 m is 200 V higher, because the field points from high to low potential.
(b) Any two of: no work is done in moving a charge along an equipotential surface; the field is always perpendicular to the surface; two equipotential surfaces never intersect; they are closer together where the field is stronger.
Shape here: flat planes perpendicular to the x axis (that is, parallel to the y–z plane), equally spaced, since the field is uniform. For example, consecutive 100 V surfaces would be 0.20 m apart, since 100/500 = 0.20 m.
Kaizen: aim for one more correct question than yesterday. That’s it. One more.
If several of those felt shaky, do not panic and do not restart the whole chapter. Find the one section the question came from, reread it, redo its worked examples with the solutions covered, and then come back. Most students who struggle with this chapter are not confused about physics — they are rushing the unit conversions and the signs. Slow those two things down and the marks arrive.