Potential due to a System of Charges

Superposition

For charges $q_1,q_2,\dots,q_n$ at distances $r_{1P},r_{2P},\dots,r_{nP}$ from the point $P$, each contributes independently and the potentials add as scalars:

$$V=V_1+V_2+\cdots+V_n=\dfrac{1}{4\pi\varepsilon_0}\left(\dfrac{q_1}{r_{1P}}+\dfrac{q_2}{r_{2P}}+\cdots+\dfrac{q_n}{r_{nP}}\right)$$

• Signs of the charges are carried into the sum. There is no angle in this formula — that is the whole advantage of potential over field.
• For a continuous distribution of charge density $\rho$, divide it into elements $\Delta v$ carrying $\rho\,\Delta v$, write the potential of each, and integrate over the distribution.

Uniformly charged spherical shell (or charged conductor)

Outside, the shell behaves as if all the charge sat at the centre:

$$V=\dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{r}\qquad(r\ge R)$$

Inside, $E=0$, so no work is done in moving a charge about; the potential is constant and equal to its surface value:

$$V=\dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{R}\qquad(r\le R)$$

• $E$ jumps discontinuously across the surface (from $0$ to $\dfrac{\sigma}{\varepsilon_0}$); $V$ is continuous everywhere.
• Corollary used in a board answer: the work done in moving any charge between two points on the surface of a charged sphere is zero, since the surface is an equipotential.

Point of zero potential on a line of charges

Set the algebraic sum to zero and solve for the position. For $q$ at $A$ and $-2q$ at $B$, separated by $d$, taking $P$ between them at distance $x$ from $A$:

$$\dfrac{kq}{x}=\dfrac{2kq}{d-x}\ \Rightarrow\ d-x=2x\ \Rightarrow\ x=\dfrac{d}{3}$$

• So the potential vanishes at $\dfrac{d}{3}$ from $q$, between the charges. Searching outside the pair, on the side of the smaller charge, gives the second solution. For $3\times10^{-8}\ \text{C}$ and $-2\times10^{-8}\ \text{C}$ placed $15\ \text{cm}$ apart, the two zeros lie at $9\ \text{cm}$ and $45\ \text{cm}$ from the positive charge.
• Always state that the zero of potential has been taken at infinity — the formula depends on it.

Two conducting balls joined by a wire

Balls of radii $r_1$ and $r_2$ carry $q_1$ and $q_2$. Joining them by a wire lets charge flow until both are at the same potential; total charge $q_1+q_2$ is conserved and spreads over the combined area $4\pi(r_1^{2}+r_2^{2})$ with a common surface density

$$\sigma=\dfrac{q_1+q_2}{4\pi\left(r_1^{2}+r_2^{2}\right)}$$

$$q_1'=\sigma\cdot4\pi r_1^{2}=\dfrac{(q_1+q_2)r_1^{2}}{r_1^{2}+r_2^{2}},\qquad q_2'=\sigma\cdot4\pi r_2^{2}=\dfrac{(q_1+q_2)r_2^{2}}{r_1^{2}+r_2^{2}}$$

Coalescing droplets

$N$ identical drops of radius $r$, each carrying $q$ and at potential $V=\dfrac{kq}{r}$, merge without loss of charge.

• Volume is conserved: $\dfrac43\pi R^{3}=N\cdot\dfrac43\pi r^{3}\Rightarrow R=N^{1/3}r$.
• Charge adds: $Q=Nq$.

$$V'=\dfrac{1}{4\pi\varepsilon_0}\dfrac{Nq}{N^{1/3}r}=N^{2/3}\,V$$

Diagram to describe in words

A point $P$ with several charges $q_1,q_2,q_3$ scattered around it, each joined to $P$ by a straight line labelled $r_{1P}$, $r_{2P}$, $r_{3P}$, and a note that the contributions add as numbers with signs. Alongside, a shell of radius $R$ with a $V$ against $r$ graph: a horizontal line at $\dfrac{kq}{R}$ up to $r=R$, then a $1/r$ tail.

Standard results