Electrostatic Potential & Capacitance

Scalar — potential due to several charges is a plain algebraic sum, $V = \sum_i kq_i/r_i$, no vector addition.

• Every point on the surface has the same $V$; no work done moving a charge along it.
• The field $\mathbf{E}$ is always perpendicular to an equipotential surface.
• Point charge → concentric spheres; uniform field → equally spaced flat planes.

For a uniform field between parallel plates: $E = V/d$. Field points toward decreasing potential; a large field needs $V$ to change rapidly with position, not just be large.

Minimum (most stable) at $\theta=0$; maximum at $\theta=180^\circ$.

• Field inside the bulk is exactly zero.
• Entire conductor (surface included) is one equipotential.
• Net charge resides only on the outer surface.
• Just outside the surface: $E = \sigma/\varepsilon_0$, perpendicular to it.

$\varepsilon_0 \approx 8.85\times10^{-12}\text{ F/m}$. Depends only on geometry ($A$, $d$), never on $Q$ or $V$.

What stays fixed as plate separation $d$ changes:

1. Connected ($V$ fixed by battery): as $d$ grows, $C$ falls, so $Q=CV$, $E=V/d$, and $U$ all fall together.
2. Disconnected ($Q$ trapped): $E=Q/(\varepsilon_0 A)$ depends only on fixed charge density, so E cannot change as $d$ grows. But $V=Ed$ climbs in direct proportion, and so does $U=Q^2/2C$.

See it live: interactive capacitor simulation →

Dielectric constant $K>1$ always increases capacitance, by weakening the net field inside via induced polarisation charges.

1. Series (same $Q$, voltages add): $\dfrac{1}{C_s}=\dfrac{1}{C_1}+\dfrac{1}{C_2}+\cdots$ — always smaller than the smallest $C$.
2. Parallel (same $V$, charges add): $C_p=C_1+C_2+\cdots$ — always larger than the largest $C$.

Factor of $\tfrac12$ because charging pushes $Q$ against a growing opposing voltage — the area under the Q–V graph, not simply $QV$.