Chapter 2: Electrostatic Potential and Capacitance

Chapter 1 introduced the vector description of electrostatics — the field $\vec{E}$. This chapter develops the scalar description — the electrostatic potential $V$. Because $V$ is a scalar (one number per point, not three), it is easier to compute, easier to add, and easier to visualize via equipotential surfaces. The two descriptions are connected by $\vec{E} = -\nabla V$.

The chapter then applies these ideas to capacitors — devices that store charge and energy in an electric field. Capacitors are everywhere: camera flashes, defibrillators, computer DRAM, radio tuners.

Concept Map

• 2.1 Electrostatic potential — definition and sign convention
• 2.2 Potential due to a point charge — derivation
• 2.3 Potential due to a system of charges and a dipole
• 2.4 Equipotential surfaces — properties
• 2.5 Relation between $E$ and $V$ — gradient form
• 2.6 Electrostatic potential energy — pair, system, dipole in field
• 2.7 Conductors in electrostatic equilibrium
• 2.8 Electrostatic shielding
• 2.9 Dielectrics — polarization, dielectric constant $K$
• 2.10 Capacitors — $Q = CV$
• 2.11 Parallel plate capacitor — $C = \varepsilon_0 A/d$; with dielectric
• 2.12 Combination of capacitors — series and parallel
• 2.13 Energy stored — derivation; energy density
• 2.14 Dielectric inserted with battery (V const) vs disconnected (Q const)

2.1 Electrostatic Potential

Definition

The electrostatic potential $V$ at a point $P$ is the work done by an external agent (against the electrostatic force) in bringing a unit positive test charge from infinity (the conventional reference) to $P$, quasi-statically:

Equivalently, in terms of the field:

Units. $1\text{ volt} = 1\text{ J/C}$. Also $\text{V/m}$ for field.

Sign Convention

• $V > 0$ near a positive charge (positive work done against the field to bring a positive test charge close).
• $V < 0$ near a negative charge.
• $V(\infty)=0$ by convention.

Relation to Potential Energy

If a charge $q$ is placed where the potential is $V$, the potential energy of the charge in this field is

The change in PE moving from $A$ to $B$ is $\Delta U = q(V_B - V_A) = -W_{\text{field}}$.

Worked Example

A test charge $q_0 = +1\text{ nC}$ is moved from $A$ to $B$ where $V_A = 100\text{ V}$ and $V_B = 30\text{ V}$. Work done by the electric field:

The field does positive work moving a $+$ charge from higher to lower potential.

Pitfalls

• $V$ is a property of the field/source, defined at every point — the test charge is conceptual only.
• The sign of work: $W_{\text{ext}} = -W_{\text{field}} = q(V_B - V_A)$.
• Choice of reference is arbitrary; only differences matter.

2.2 Potential Due to a Point Charge

Definition / Derivation

Bring a unit positive test charge from $\infty$ to a distance $r$ from a point charge $q$, along a radial path.

Step 1. Field at distance $r'$: $E = kq/r'^2$, radially outward (for $q>0$).

Step 2. Work done by the field when test charge moves from $r'$ to $r'-dr'$ (toward $q$): $dW_{\text{field}} = -E\,dr' \cdot(\text{sign of motion})$. Going from $\infty$ to $r$, $dl = -dr'$ along outward $\hat{r}$, so $\vec{E}\cdot d\vec{l} = E\,(-dr')$:

Step 3.

Notes:

• For $q>0$, $V>0$ and decreases as $1/r$.
• For $q< 0$, $V< 0$, and increases toward zero as $r\to\infty$.

Worked Example

A charge $+5\,\mu\text{C}$ is at the origin. The potential at $r = 0.3\text{ m}$:

Pitfalls

• $V \propto 1/r$, not $1/r^2$ — that's the field.
• $V$ remains finite at $r$ even where $\vec{E}$ may seem strong (so long as you don't sit on the point charge).

2.3 Potential Due to Systems & Dipoles

System of Point Charges (Superposition)

Because $V$ is a scalar, the potential at $P$ due to a collection of point charges is the algebraic sum:

Derivation — Dipole Potential at a General Point

Place $-q$ at $-\vec{a}$ and $+q$ at $+\vec{a}$. Let $P$ be at $\vec{r}$, with $r\gg a$. Angle between $\vec{r}$ and $\vec{p}$ is $\theta$.

Step 1. Distance from $+q$ to $P$: $r_+ = r - a\cos\theta$ (to first order). Distance from $-q$ to $P$: $r_- = r + a\cos\theta$.

Step 2.

Step 3. For $r\gg a$, $r^2 - a^2\cos^2\theta \approx r^2$, and $p = q\cdot 2a$:

Special Points

• Axial ($\theta=0$): $V = kp/r^2$.
• Equatorial ($\theta=\pi/2$): $V = 0$ (any point equidistant from $+$ and $-$ has zero net potential).
• Behind the axis ($\theta=\pi$): $V = -kp/r^2$.

Worked Example

Dipole with $p = 4\times 10^{-9}\text{ C m}$. Find $V$ at $r=0.2\text{ m}$, $\theta = 60°$.

Pitfalls

• Dipole potential falls as $1/r^2$ (faster than point-charge $1/r$). Dipole field falls as $1/r^3$.
• Equatorial potential is zero, but the field there is not zero.

2.4 Equipotential Surfaces

Definition

An equipotential surface is one on which $V$ has the same value at every point.

Properties

1. Work done by the field to move a charge along an equipotential is zero ($\Delta V = 0$).
2. $\vec{E}$ is always perpendicular to equipotentials. (If $\vec{E}$ had a component along the surface, moving a charge along that direction would change $V$.)
3. Equipotentials never intersect (at an intersection $V$ would have two values).
4. Closely spaced equipotentials indicate strong $\vec{E}$; widely spaced ones indicate weak $\vec{E}$.
5. Inside a conductor in equilibrium, the entire volume is at one potential — the conductor is an equipotential.

Standard Patterns

• Point charge: concentric spheres centered on the charge.
• Uniform field: parallel planes perpendicular to $\vec{E}$.
• Dipole: complicated curves — except the equatorial plane, which is the $V=0$ equipotential.

Worked Example

A uniform field $\vec{E} = 100\,\hat{i}\text{ V/m}$ exists in a region. Sketch the equipotentials passing through $x=0,1,2,3\text{ m}$.

$V(x) = -\int_0^x E\,dx' = -100x$ (taking $V(0)=0$). Equipotentials at $0, -100, -200, -300\text{ V}$ are vertical planes spaced $1\text{ m}$ apart.

Pitfalls

• Equipotential $\neq$ equal-field surface. A point at distance $r$ from $+q$ is at the same potential as a point at distance $r$ from $-q$ on the equatorial plane (both zero) — but the fields can be very different.
• Equipotentials need not be closed surfaces.

2.5 Relation Between $E$ and $V$

Derivation

For a small displacement $d\vec{l}$, the change in potential is

where $E_l$ is the component of $\vec{E}$ along $d\vec{l}$.

Step 1. Choosing $d\vec{l}$ along the direction of fastest decrease of $V$ gives

Step 2. In Cartesian form,

The minus sign: $\vec{E}$ points from high $V$ to low $V$.

Worked Example

Given $V(x,y,z) = 5x^2 - 3y + 2z$ (in volts, $x,y,z$ in meters), find $\vec{E}$.

At the origin: $\vec{E} = 3\hat{j} - 2\hat{k}\text{ V/m}$.

Pitfalls

• $V$ is constant inside a conductor $\Rightarrow$ $\vec{E}=0$ inside (consistent with electrostatic equilibrium).
• The minus sign is essential — students often drop it.
• For spherically symmetric $V(r)$: $\vec{E} = -(dV/dr)\hat{r}$.

2.6 Electrostatic Potential Energy

(a) Two Point Charges

The PE of two point charges $q_1, q_2$ separated by $r$ is the work done to bring them from infinity to that configuration:

Sign: positive for like charges (positive work needed), negative for unlike (energy released).

(b) System of $n$ Charges

Sum over all distinct pairs:

Worked Example — Three Charges

Charges $+q, +q, +q$ at vertices of an equilateral triangle of side $a$.

For three charges $+q, -q, +q$ at the same vertices:

(c) Dipole in an External Field

(Derived in 1.8.) With reference $U(\theta = \pi/2) = 0$:

(d) Worked Example — Energy released

Take two charges $+q$ and $-q$ from separation $2r$ to separation $r$. $\Delta U = (-kq^2/r) - (-kq^2/(2r)) = -kq^2/(2r)$. Energy released $= +kq^2/(2r)$ (system more bound).

Pitfalls

• $U$ is a property of the configuration, not of a single charge.
• For pairs, do not double-count: $\sum_{i<j}$, not $\sum_{i\neq j}$.
• Sign of $U$: same-sign charges $\to U > 0$; unlike $\to U < 0$.

2.7 Conductors in Electrostatic Equilibrium

Definition

A conductor contains free electrons. In electrostatic equilibrium these have rearranged so that no net flow occurs.

Key Properties

1. $\vec{E}=0$ inside the conductor (otherwise free charges would still move).
2. $V$ is constant throughout the conductor (since $\vec{E}=0 \Rightarrow \nabla V = 0$).
3. Net charge resides on the outer surface (Gauss's law inside: enclosed charge $=0$).
4. $\vec{E}$ at the surface is perpendicular to the surface (any tangential component would drive currents).
5. Magnitude at the surface: $E = \sigma/\varepsilon_0$.
6. Surface charge density is higher where curvature is higher (sharp points $\to$ high $\sigma$ $\to$ corona discharge).

Derivation — $E = \sigma/\varepsilon_0$ just outside

Use a small Gaussian pillbox straddling the surface, faces parallel to the surface.

• Field inside conductor: $0$. So flux through inner face is $0$.
• Field outside is perpendicular to surface (property 4). Flux through outer face: $EA$.
• Lateral flux: zero (in the limit of thin box).
• Enclosed charge: $\sigma A$.

Gauss: $EA = \sigma A/\varepsilon_0$, so $E = \sigma/\varepsilon_0$.

Worked Example

An isolated conducting sphere of radius $R$ carries charge $Q$. Find $V$ everywhere.

• Outside ($r>R$): $V = kQ/r$ (behaves like point charge at center).
• Surface ($r=R$): $V_R = kQ/R$.
• Inside ($r<R$): $V = V_R = kQ/R$ (constant; field is zero inside).

Pitfalls

• $V$ inside is not zero — it's whatever value the surface holds.
• Field changes discontinuously at the surface (from $0$ inside to $\sigma/\varepsilon_0$ outside) but $V$ is continuous.

2.8 Electrostatic Shielding

Concept

If we hollow out a region inside a conductor (a cavity, with no charges inside), then $\vec{E} = 0$ throughout the cavity — regardless of charges outside the conductor.

Why? The conductor rearranges its surface charges to ensure $\vec{E}=0$ in its bulk; by uniqueness, the field in the cavity must also be zero (Laplace's equation with zero boundary).

Applications

• Faraday cage: electronic equipment, MRI rooms, cars during lightning.
• Coaxial cables with grounded shielding.
• Sensitive instruments shielded by metal enclosures.

Pitfalls

• Shielding works only against external electrostatic fields. A charge inside the cavity does produce a field there.
• The cage need not be solid metal — a fine mesh works for low-frequency fields (this is electromagnetic shielding territory, but the principle is electrostatic).

2.9 Dielectrics

Definition

A dielectric is an insulator with no (or few) free charges, but whose molecules can be polarized by an external field.

Polar vs Non-polar

• Non-polar molecules (e.g., $\text{H}_2, \text{O}_2$): no permanent dipole; an external field separates the positive and negative centers a tiny amount, induced dipole.
• Polar molecules (e.g., $\text{H}_2\text{O, HCl}$): have permanent dipole moment, randomly oriented in absence of field; an external field aligns them partially.

In both cases, the result is a polarization $\vec{P}$ — dipole moment per unit volume.

Dielectric Constant

When a dielectric fills a capacitor, the induced polarization reduces the net field. The dielectric constant $K$ (also $\varepsilon_r$) is:

where subscript $0$ denotes vacuum value. $K \ge 1$ always; $K_{\text{vacuum}} = 1$; $K_{\text{air}} \approx 1.0006$; $K_{\text{water}} \approx 80$; $K_{\text{mica}}\approx 6$.

Worked Example

A parallel plate capacitor in vacuum has $E_0 = 10^5\text{ V/m}$. A slab with $K=4$ is inserted (fully). The new field inside the slab is

(Outside the slab, $E$ is unchanged.)

Pitfalls

• A dielectric reduces $E$ inside it but does not eliminate it (unlike a conductor).
• A conductor is the $K\to\infty$ limit, in the sense that the field inside is reduced to zero.

2.10 Capacitors

Definition

A capacitor is any pair of conductors separated by an insulator, used to store charge $\pm Q$ on the two plates. The plates are at potentials $V_+$ and $V_-$.

The capacitance is

Unit. Farad ($\text{F} = \text{C/V}$). Practical units: $\mu\text{F}$, $\text{nF}$, $\text{pF}$.

$C$ depends only on the geometry and the dielectric between the plates, not on $Q$ or $V$.

2.11 Parallel Plate Capacitor

Derivation — $C$ in Vacuum

Two large parallel plates, area $A$, separation $d$, charges $\pm Q$.

Step 1. Surface densities: $\sigma = Q/A$ on the $+$ plate, $-\sigma$ on the $-$ plate.

Step 2. Between the plates (treating them as infinite sheets): $E = \sigma/\varepsilon_0$.

Step 3. Potential difference: $V = E\cdot d = \sigma d/\varepsilon_0 = Qd/(\varepsilon_0 A)$.

Step 4. $C = Q/V$:

With Dielectric Fully Filling the Gap

Field reduced by $K$: $E = \sigma/(K\varepsilon_0)$, so $V = Ed = Qd/(K\varepsilon_0 A)$.

Partial Dielectric (slab thickness $t<d$)

By integration of $E$ over the gap:

For $K\to\infty$ (conducting slab): $C \to \varepsilon_0 A/(d-t)$ — effective plate separation reduced by $t$.

Worked Example

A parallel plate capacitor: $A = 100\text{ cm}^2 = 10^{-2}\text{ m}^2$, $d = 1\text{ mm} = 10^{-3}\text{ m}$. In vacuum:

With mica ($K=6$) filling: $C = 6\times 88.5 = 531\text{ pF}$.

Pitfalls

• Edge effects (fringing fields) are ignored — true only for $d\ll \sqrt{A}$.
• $C$ depends only on geometry and dielectric, not on charge.
• If a conductor is inserted (not dielectric), $C \to \varepsilon_0 A/(d-t)$ where $t$ is the conductor's thickness.

2.12 Combination of Capacitors

Parallel

Capacitors in parallel share the same potential difference $V$.

$Q = Q_1 + Q_2 + \cdots = (C_1 + C_2 + \cdots)V$.

Series

Capacitors in series carry the same charge $Q$ (induced charges).

$V = V_1 + V_2 + \cdots = Q/C_1 + Q/C_2 + \cdots = Q\left(\frac{1}{C_1}+\frac{1}{C_2}+\cdots\right)$.

For two capacitors in series: $C = C_1 C_2/(C_1+C_2)$.

Worked Example

$C_1 = 2\,\mu\text{F}$, $C_2 = 3\,\mu\text{F}$, $C_3 = 6\,\mu\text{F}$.

• All three in parallel: $C = 11\,\mu\text{F}$.
• All three in series: $1/C = 1/2 + 1/3 + 1/6 = 1 \Rightarrow C = 1\,\mu\text{F}$.

If $50\text{ V}$ is applied:

• Parallel: $Q_1 = 100\,\mu\text{C}$, $Q_2 = 150\,\mu\text{C}$, $Q_3 = 300\,\mu\text{C}$.
• Series: $Q = CV = 50\,\mu\text{C}$ on each. Voltages: $V_1 = 25, V_2 = 50/3, V_3 = 25/3$ (sum $=50\text{ V}$).

Pitfalls

• "Same charge" in series — even on capacitors of different capacitance.
• "Same voltage" in parallel — even when capacitors store very different charges.
• The smaller capacitor takes the larger voltage in series.

2.13 Energy Stored in a Capacitor

Derivation

Bring infinitesimal charge $dq$ from $-$ plate to $+$ plate against current potential $V'=q'/C$.

Step 1. Work done: $dW = V'\,dq' = (q'/C)\,dq'$.

Step 2. Total work to charge from $0$ to $Q$:

Step 3. Using $Q = CV$:

Energy Density

For a parallel plate capacitor, $U = \tfrac{1}{2}CV^2 = \tfrac{1}{2}(\varepsilon_0 A/d)(Ed)^2 = \tfrac{1}{2}\varepsilon_0 E^2\cdot (Ad)$. Volume of field region: $Ad$. So

This is a general result: the electric field carries energy density $\tfrac{1}{2}\varepsilon_0 E^2$ everywhere.

Worked Example

A $10\,\mu\text{F}$ capacitor charged to $100\text{ V}$:

Pitfalls

• The factor of $1/2$ comes from the averaging (you start at $V=0$ and end at $V$).
• When connecting a charged capacitor in parallel with an uncharged one, charge is conserved, but energy is not — some is dissipated as heat/radiation regardless of how slowly you connect them.

2.14 Effect of Inserting a Dielectric

Case 1 — Battery Connected (V constant)

The battery holds $V$ fixed at $V_0$.

Energy increases; extra energy comes from the battery (battery supplies $\Delta Q$ at $V_0$, doing work $V_0\,\Delta Q = V_0(KC_0V_0 - C_0V_0) = (K-1)C_0V_0^2$; half goes to extra capacitor energy, half is dissipated).

Case 2 — Battery Disconnected (Q constant)

Energy decreases; work is done by the field on the dielectric as it is drawn in.

Worked Example

A capacitor with $C_0 = 1\,\mu\text{F}$ is charged to $100\text{ V}$ then disconnected. A slab with $K=4$ is inserted.

• $Q_0 = 10^{-4}\text{ C}$ unchanged.
• New $C = 4\,\mu\text{F}$.
• New $V = 25\text{ V}$.
• New $U = \tfrac{1}{2}(4\times 10^{-6})(25)^2 = 1.25\times 10^{-3}\text{ J}$ (decreased from $5\times 10^{-3}\text{ J}$).
• Energy lost = $3.75\text{ mJ}$, done as work on the dielectric (or radiated).

Pitfalls

• The two scenarios give opposite changes in energy — read the problem carefully.
• $E$ inside the dielectric changes in both cases: in case 1 it stays the same; in case 2 it drops.

Solved Problems

Problem 1 (Easy)

Find the work done in moving a $+2\,\mu\text{C}$ charge from a point where $V = -50\text{ V}$ to a point where $V = +200\text{ V}$.

Problem 2 (Easy)

Three capacitors of $1, 2, 3\,\mu\text{F}$ are connected in series across $11\text{ V}$. Find the charge and voltage across each.

$1/C = 1 + 1/2 + 1/3 = 11/6 \Rightarrow C = 6/11\,\mu\text{F}$. $Q = CV = (6/11)\times 11 = 6\,\mu\text{C}$ on each. $V_1 = 6\text{ V}, V_2 = 3\text{ V}, V_3 = 2\text{ V}$ (sum $= 11\text{ V}$).

Problem 3 (Medium)

Two capacitors of $C_1 = 3\,\mu\text{F}$ (charged to $300\text{ V}$) and $C_2 = 2\,\mu\text{F}$ (uncharged) are connected together (positive to positive). Find the common potential and energy loss.

Charge conservation: $Q_1 + Q_2 = C_1\cdot 300 = 900\,\mu\text{C}$. Common potential: $V = 900/(C_1+C_2) = 900/5 = 180\text{ V}$. $U_{\text{before}} = \tfrac{1}{2}(3\times 10^{-6})(300)^2 = 0.135\text{ J}$. $U_{\text{after}} = \tfrac{1}{2}(5\times 10^{-6})(180)^2 = 0.081\text{ J}$. $\Delta U = 0.054\text{ J}$ dissipated.

Problem 4 (Medium)

A spherical conductor of radius $R_1 = 5\text{ cm}$, carrying charge $Q_1 = 1\,\mu\text{C}$, is connected by a thin wire to another isolated sphere of radius $R_2 = 10\text{ cm}$, initially uncharged. Find the final charges.

When connected, they reach the same potential. $V = kQ_1'/R_1 = kQ_2'/R_2 \Rightarrow Q_1'/Q_2' = R_1/R_2 = 1/2$. $Q_1' + Q_2' = 1\,\mu\text{C}$. $Q_1' = 1/3\,\mu\text{C}$, $Q_2' = 2/3\,\mu\text{C}$.

Surface densities: $\sigma_1/\sigma_2 = (Q_1'/4\pi R_1^2)/(Q_2'/4\pi R_2^2) = (1/3)(R_2^2/R_1^2)/(2/3) = R_2^2/(2R_1^2) = 100/(2\cdot 25) = 2$. So smaller sphere has higher $\sigma$ — explains lightning rods.

Problem 5 (Medium)

A parallel plate capacitor of $C = 5\,\mu\text{F}$, plate separation $4\text{ mm}$, is connected to $200\text{ V}$. A dielectric slab of $K=4$ and thickness $2\text{ mm}$ is inserted (filling half the gap). Find new $C$, $Q$, $V$, $U$.

Compared to $C_0 = \varepsilon_0 A/d = \varepsilon_0 A/4$: $C_{\text{new}}/C_0 = 4/2.5 = 1.6$. So $C_{\text{new}} = 8\,\mu\text{F}$.

If battery is still connected: $V = 200\text{ V}$, $Q = 1.6\text{ mC}$, $U = 0.16\text{ J}$ (up from $0.1\text{ J}$).

Problem 6 (Hard)

A charged isolated soap bubble of radius $R$ at potential $V$ has charge $Q = RV/k$. If it bursts into a single drop, find the potential of the drop. Assume same total charge, mass conserved (so volume conserved, surface area not).

Soap bubble: treat the surface as a thin spherical shell of charge $Q$. Inside the bubble (before bursting) the field is zero and potential is $V$.

When the bubble bursts and collapses into a single drop: the total volume of soap film, $4\pi R^2\cdot t$, becomes the volume of the drop $\tfrac{4}{3}\pi r^3$. So $r = (3R^2 t)^{1/3}$, very small.

New potential: $V' = kQ/r = V\cdot R/r = V(R/r)$, which is much larger than $V$.

(In some texts the question is "$n$ identical droplets coalesce" — see Edge Cases.)

Problem 7 (Hard)

A capacitor $C_1 = 2\,\mu\text{F}$ is charged to $V_0 = 100\text{ V}$. It is then connected through a resistor to an uncharged $C_2 = 3\,\mu\text{F}$. Find the final charges and total energy dissipated.

$Q_{\text{initial}} = 200\,\mu\text{C}$. After connection (equilibrium, same voltage): $V_f = 200/(C_1+C_2) = 200/5 = 40\text{ V}$. $Q_1' = 80\,\mu\text{C}$, $Q_2' = 120\,\mu\text{C}$. $U_i = \tfrac{1}{2}(2\times 10^{-6})(100)^2 = 0.01\text{ J} = 10\text{ mJ}$. $U_f = \tfrac{1}{2}(5\times 10^{-6})(40)^2 = 4\text{ mJ}$. Energy dissipated = $6\text{ mJ}$.

(Notably, the energy dissipated does not depend on $R$.)

JEE/NEET Edge Cases

1. Energy lost is independent of the resistor when connecting capacitors: dissipation is $\frac{C_1 C_2 (V_1-V_2)^2}{2(C_1+C_2)}$.

2. $n$ identical droplets coalesce into a big drop: big drop's radius is $n^{1/3}r$. Charge $= nq$. Potential $V_{\text{big}} = k(nq)/(n^{1/3}r) = n^{2/3} V_{\text{small}}$.

3. Capacitor with conductor slab of thickness $t$: $C = \varepsilon_0 A/(d-t)$. The slab can be anywhere between the plates; it does not matter.

4. Capacitance of an isolated sphere of radius $R$ in vacuum: $C = 4\pi\varepsilon_0 R$ (taking $V$ relative to infinity).

5. Spherical capacitor (inner radius $a$, outer $b$): $C = 4\pi\varepsilon_0 ab/(b-a)$.

6. Cylindrical capacitor (length $L$, radii $a, b$): $C = 2\pi\varepsilon_0 L/\ln(b/a)$.

7. Force between capacitor plates (battery disconnected, charge constant): $F = Q^2/(2\varepsilon_0 A) = \sigma^2 A/(2\varepsilon_0)$ — half of what you'd expect from "naively" applying $F = qE$ because each plate sees only the other plate's field, $\sigma/(2\varepsilon_0)$, not the full $\sigma/\varepsilon_0$.

8. Maximum voltage before dielectric breakdown: $V_{\max} = E_{\text{br}}\cdot d$. Different dielectrics: air $\sim 3\times 10^6\text{ V/m}$, mica $\sim 10^8\text{ V/m}$.

9. Combination of three capacitors at corners of a triangle with one missing edge — often a Wheatstone-bridge-like situation; if balance condition holds, the bridge capacitor carries no charge.

10. Inside a hollow conductor at potential $V_0$: $V = V_0$ everywhere inside, even if there is a cavity with no charge.

Quick Recap

• $V(P) = -\int_\infty^P \vec{E}\cdot d\vec{l}$; $V = kq/r$ for point charge.
• $\vec{E} = -\nabla V$; perpendicular to equipotentials.
• $U_{12} = kq_1q_2/r$; $U_{\text{dipole}} = -\vec{p}\cdot\vec{E}$.
• Conductors: $\vec{E}=0$ inside, $V$ constant, $E = \sigma/\varepsilon_0$ just outside.
• $C = Q/V$; parallel plate $C_0 = \varepsilon_0 A/d$.
• Series: $1/C_{\text{eq}} = \sum 1/C_i$. Parallel: $C_{\text{eq}} = \sum C_i$.
• $U = \tfrac{1}{2}CV^2 = \tfrac{1}{2}Q^2/C$; energy density $u = \tfrac{1}{2}\varepsilon_0 E^2$.
• Dielectric (battery connected): $V$ const, $Q$ and $U$ rise by $K$.
• Dielectric (disconnected): $Q$ const, $V$ and $U$ drop by $K$.

Formula Sheet