Electrostatic Potential & Capacitance: Full Notes

2.1 Electrostatic Potential

Just as a mass raised against gravity stores potential energy, moving a charge against an electric field stores electrostatic potential energy. The potential at a point is this stored energy per unit charge, the work needed to bring a tiny positive test charge from infinity (where potential is conventionally zero) to that point, without speeding it up:

V(r) = \dfrac{W_{\infty \to r}}{q_0}

For a single point charge $Q$ , carrying out this integral of the Coulomb field gives a strikingly simple result:

Potential of a point charge

\textcolor{#e08a1e}{V(r)} = \dfrac{1}{4\pi\varepsilon_0}\dfrac{Q}{r} = \dfrac{kQ}{r}

Unlike the electric field, potential is a scalar, so the potential due to several charges is just an ordinary sum, no vector addition required:

V = \sum_i \dfrac{kq_i}{r_i}

2.2 Equipotential Surfaces

An equipotential surface is any surface on which every point has the same potential. For a single point charge, these are concentric spheres centred on the charge. Two properties worth holding onto:

No work is done moving a charge along an equipotential surface, since $W = q\Delta V$ and $\Delta V = 0$ there.
The electric field is always perpendicular to an equipotential surface at every point. If it had any component along the surface, moving charge along that direction would require zero work against a nonzero force component, a contradiction.

2.3 Relation Between Field and Potential

Field and potential are two descriptions of the same underlying reality, one a vector, one a scalar, and they’re linked by a derivative. Moving a tiny distance $dr$ against the field changes the potential by $dV = -E\,dr$ , so:

Field from potential

\textcolor{#e08a1e}{E} = -\dfrac{dV}{dr}

The minus sign matters: the field always points in the direction of decreasing potential, “downhill”, the fastest. A large field doesn’t require a large potential; it requires potential to be changing rapidly with position. A region of uniformly high potential with no variation at all has zero field inside it.

Worked example

Field between two charged plates, from potential alone

Two large parallel plates are held at 0 V and 200 V, 4 mm apart. Find the field between them.

E = \dfrac{\Delta V}{d} = \dfrac{200}{0.004} = \textcolor{#e08a1e}{50{,}000\text{ V/m}}

A uniform field between parallel plates is exactly $E = V/d$ , the straight-line version of $E=-dV/dr$ since the potential changes at a constant rate across the gap.

2.4 Electrostatic Potential Energy

The potential energy of a pair of charges is the work done to assemble them, bringing the second charge in from infinity against the first charge’s field:

U = \dfrac{kq_1q_2}{r_{12}}

For an electric dipole (moment $p$ ) sitting in an external uniform field $E$ at angle $\theta$ to it, the potential energy is:

U(\theta) = -pE\cos\theta = -\vec{p}\cdot\vec{E}

which is minimum (most stable) when the dipole aligns with the field, and maximum when it points directly against it, exactly the torque $\tau = pE\sin\theta$ trying to rotate it toward.

2.5 Conductors in Electrostatic Equilibrium

Inside a conductor in electrostatic equilibrium, the field is always exactly zero (free charges would otherwise keep moving until it is). Three consequences follow immediately:

The entire conductor, surface included, is one equipotential.
Any net charge resides entirely on the outer surface, never in the bulk.
Just outside the surface, the field is perpendicular to it, with magnitude $E = \sigma/\varepsilon_0$ .

2.6 Capacitors and Capacitance

A capacitor is any pair of conductors that can store equal and opposite charges, separated by an insulator. Experimentally, the charge stored is always directly proportional to the potential difference applied, and the constant of proportionality, which depends only on the geometry, is called the capacitance:

Capacitance

\textcolor{#e08a1e}{C} = \dfrac{Q}{V}

with SI unit the farad (F), 1 F = 1 C/V, an enormous unit in practice, real capacitors are usually rated in microfarads (µF) or picofarads (pF).

2.7 The Parallel-Plate Capacitor

For two large plates of area $A$ , separated by a small gap $d$ , carrying charges $+Q$ and $-Q$ , Gauss’s law gives a uniform field between them, $E = \sigma/\varepsilon_0 = Q/(\varepsilon_0 A)$ . Integrating $V = Ed$ across the gap and substituting into $C=Q/V$ :

Parallel-plate capacitance

C = \dfrac{\varepsilon_0 A}{d}

Capacitance depends only on the plates’ area and separation, never on how much charge happens to be on them, exactly the same way a cup’s volume doesn’t depend on how much water is currently in it.

2.7.1 The connected-vs-disconnected distinction

What happens when $d$ changes depends entirely on whether a battery is still attached:

Battery connected: $V$ is held fixed by the battery. As $d$ grows, $C$ falls, so $Q=CV$ falls too (charge flows back through the battery), and with it $E=V/d$ and the stored energy.
Battery disconnected: $Q$ is trapped, fixed. Since $E=Q/(\varepsilon_0 A)$ depends only on the charge density, the field cannot change as $d$ grows. But $V=Ed$ with $E$ fixed and $d$ growing means $V$ climbs in direct proportion. The concept page’s simulation shows exactly this, see the interactive capacitor for the live version.

2.8 Dielectrics

Filling the gap between the plates with an insulating material (a dielectric) always increases capacitance, by a factor called the dielectric constant $K$ (also written $\varepsilon_r$ ):

C_{dielectric} = K\varepsilon_0\dfrac{A}{d} = KC_{vacuum}

The mechanism: the external field polarises the dielectric’s molecules, producing an induced field that opposes and partially cancels the original field inside the material, so a given charge now produces a smaller net field, and therefore a smaller voltage for the same charge, which by $C=Q/V$ means a larger $C$ .

2.9 Combinations of Capacitors

2.9.1 Series

Same charge $Q$ sits on every capacitor in the chain; voltages add:

\dfrac{1}{C_{series}} = \dfrac{1}{C_1}+\dfrac{1}{C_2}+\cdots

always smaller than the smallest individual capacitance, the same combination rule as resistors in parallel.

2.9.2 Parallel

Same voltage $V$ across every capacitor; charges add:

C_{parallel} = C_1+C_2+\cdots

always larger than the largest individual capacitance, more plate area effectively available to store charge.

Worked example

Two capacitors, both ways

A 2 µF and a 3 µF capacitor are combined (a) in series, (b) in parallel. Find the equivalent capacitance each way.

\text{(a)}\ \dfrac{1}{C_s} = \dfrac{1}{2}+\dfrac{1}{3} = \dfrac{5}{6} \;\Rightarrow\; C_s = \textcolor{#e08a1e}{1.2\ \mu\text{F}}

\text{(b)}\ C_p = 2+3 = \textcolor{#e08a1e}{5\ \mu\text{F}}

Series always lands below the smaller value (1.2 < 2); parallel always lands above the larger value (5 > 3).

2.10 Energy Stored in a Capacitor

Charging a capacitor means pushing charge onto a plate against an opposing voltage that itself grows as charge builds up, so the energy isn’t simply $QV$ , it’s the area under the Q-V graph, a triangle, giving a factor of $\tfrac{1}{2}$ :

Energy stored

U = \dfrac{1}{2}QV = \dfrac{1}{2}CV^2 = \dfrac{Q^2}{2C}

Worked example

Energy in a charged capacitor

A 10 µF capacitor is charged to 200 V. Find the energy stored.

U = \dfrac{1}{2}CV^2 = \dfrac{1}{2}(10\times10^{-6})(200)^2 = \textcolor{#e08a1e}{0.2\text{ J}}

— From the NCERT exercises

A classic exercise question from the chapter, with an original worked solution.

NCERT Exercise 2.8

A 600 pF capacitor is charged by a 200 V supply. It is then disconnected from the supply and connected to another uncharged 600 pF capacitor. How much electrostatic energy is lost in the process?

Solution

Initial charge and energy, on the first capacitor alone:

Q_0 = C V_0 = (600\times10^{-12})(200) = 1.2\times10^{-7}\text{ C}

U_i = \dfrac{1}{2}C V_0^2 = \dfrac{1}{2}(600\times10^{-12})(200)^2 = 1.2\times10^{-5}\text{ J}

Connecting an identical uncharged capacitor shares the same total charge equally (both have equal $C$ ), so each ends up at half the original voltage, 100 V, across double the capacitance:

U_f = \dfrac{1}{2}(2C)(100)^2 = \dfrac{1}{2}(1200\times10^{-12})(100)^2 = 6\times10^{-6}\text{ J}

\Delta U = U_i - U_f = 1.2\times10^{-5} - 0.6\times10^{-5} = \textcolor{#e08a1e}{6\times10^{-6}\text{ J lost}}

This energy doesn’t vanish, it’s dissipated as heat and a brief spark in the connecting wires, which have some resistance even though it’s never mentioned in the problem.