Electric Potential

The question this answers

An electron sits at rest inside a television tube. You switch on 120 volts across the tube, and a fraction of a nanosecond later the electron is moving at six and a half million metres per second. Nobody told you the shape of the field inside the tube, how the wires are bent, or what path the electron took. Yet the speed is fixed, and you can get it in one line.

That is what this lesson buys you. In Gauss's law you found $\vec E$, a vector at every point, and adding two fields meant adding two sets of components. Now you will trade that vector for a single number at each point in space. The number carries almost all the same information, adds like ordinary arithmetic, and plugs straight into energy conservation. Most electrostatics problems that look brutal in terms of field are short in terms of potential.

Work, energy, and the quantity that does not depend on the charge

Start with what you already know about the electric force. Push a charge $q$ from point $A$ to point $B$ through a field $\vec E$. The force on it is $q\vec E$, so the work the field does is

The electric force of a point charge is central and falls off as $1/r^2$, just like gravity, and that makes it conservative: the work depends only on the endpoints, not on the wiggly path you take between them. So you can define a potential energy $U$ with $W_{A\to B} = U_A - U_B$, exactly as you did for gravity.

Here is the useful observation. Double the charge you are pushing and the force doubles, so the work doubles, so $U$ doubles. The energy is proportional to $q$. Divide it out and what is left belongs to the field alone. That quantity is the electric potential $V$:

Written as the two statements you will actually use:

The unit of potential is the volt, one joule per coulomb. A 9 V battery means each coulomb of charge that crosses from one terminal to the other gains or loses 9 joules. That is the whole definition.

Two consequences fall straight out. First, only differences of $V$ are physical, because only differences of energy are. You are free to call any one point zero, and for a charge distribution of finite size the convenient choice is $V = 0$ infinitely far away. Second, if nothing but the electric force acts,

energy conservation with $q\Delta V$ in place of $mgh$.

Example 1. An electron starts from rest and is accelerated through 120 V. How fast is it going?

The electron has charge $q = -e = -1.6\times10^{-19}\ \text{C}$, and it speeds up, so it must move toward higher potential: $\Delta V = +120\ \text{V}$. Then

It loses that much potential energy, so it gains $1.92\times10^{-17}\ \text{J}$ of kinetic energy:

The fraction is $3.84\times10^{-17}/9.11\times10^{-31} = 4.215\times10^{13}$, and its square root is $6.49\times10^{6}\ \text{m/s}$. If the particle starts with speed already, add the initial kinetic energy first rather than using the formula for $v$ directly: $\tfrac12 m v_f^2 = \tfrac12 m v_i^2 + |q\Delta V|$.

Potential of a point charge, and why adding is easy

Put a single charge $Q$ at the origin. Its field is radial with magnitude $kQ/r^2$. Walk in from infinity along a radial line to a distance $r$. Then $\vec E\cdot d\vec\ell = E_r\,dr$ and

With the usual choice $V(\infty) = 0$,

Read it carefully. It falls as $1/r$, not $1/r^2$: one power slower than the field, because integrating a field over distance adds a power of $r$. It has a sign: a negative charge produces negative potential everywhere. And $r$ is a plain distance, never a component, so there are no angles in this formula.

That last point is what makes potential the easier tool. Potential energy is a scalar, so potentials from several charges simply add:

Four numbers added, with their signs. Compare that with adding four field vectors and resolving each into components.

Example 2. Four charges of $6.0\ \text{nC}$ each sit at the corners of a square of side $20\ \text{cm}$. Find $V$ at the centre.

First, the geometry. The centre is half a diagonal from each corner. The diagonal is $0.20\sqrt2 = 0.2828\ \text{m}$, so each distance is $r = 0.1414\ \text{m}$. All four are equal, so

The numerator is $4\times53.94 = 215.8$, and dividing by $0.1414$ gives $V = 1526\ \text{V}$. Note what did not happen: you never asked which way any of the four fields pointed. By symmetry $\vec E$ at the centre is zero, while $V$ is emphatically not. Those two facts live independently.

Continuous charge: cut it into pieces and add the numbers

For a spread-out charge, chop it into pieces $dq$ small enough to count as points, and add:

where $r$ is the distance from the piece to your field point. Two shapes cover most problems.

A ring, on its axis. A ring of radius $R$ carries total charge $Q$. Take a point on the axis a distance $z$ from the centre. Every piece of the ring, no matter where it sits around the circle, lies at the same distance $\sqrt{z^2+R^2}$ from that point. So $r$ comes out of the integral and what is left is just $\int dq = Q$:

No integration at all. For a ring of radius $10\ \text{cm}$ with $12\ \text{nC}$ at $z = 20\ \text{cm}$: $\sqrt{0.04+0.01} = 0.2236\ \text{m}$, and $V = (8.99\times10^{9})(12\times10^{-9})/0.2236 = 107.9/0.2236 = 482\ \text{V}$.

A rod, along its own axis. A rod of length $L$ carries charge per unit length $\lambda$. You stand on the line of the rod, a distance $d$ beyond its near end. A slice of length $dx$ at distance $x$ from you holds $dq = \lambda\,dx$, and it sits at distance $x$. So

With $\lambda = 3.0\ \text{nC/m}$, $L = 0.50\ \text{m}$, $d = 0.20\ \text{m}$: $k\lambda = 26.97$, and $\ln(0.70/0.20) = \ln 3.5 = 1.2528$, giving $V = 33.8\ \text{V}$. Doing this as a field integral would mean the same work plus a component argument on top.

Getting the field back out

Potential throws away nothing, because you can differentiate your way back. Move a small step $dx$ along $x$. The definition says $dV = -\vec E\cdot d\vec\ell = -E_x\,dx$, so

and likewise for $y$ and $z$ (written together, $\vec E = -\nabla V$). The field is the downhill slope of the potential. Steep $V$ means strong field; flat $V$ means no field.

Two quick applications. If $V(x) = 400 - 2500x$ volts inside a parallel-plate gap, then $E_x = -(-2500) = +2500\ \text{V/m}$: uniform, as the straight-line $V$ promised. If $V(x) = 60x^2 - 15x$, then $E_x = -(120x - 15)$, which at $x = 0.25\ \text{m}$ is $-(30-15) = -15\ \text{V/m}$. Volts per metre and newtons per coulomb are the same unit.

Running it the other way, a uniform field $E$ along $x$ gives $V_A - V_B = E(x_B - x_A)$. For $E = 1800\ \text{N/C}$ with $A$ at $0.10$ m and $B$ at $0.55$ m: $V_A - V_B = 1800(0.45) = 810\ \text{V}$. Potential drops as you move along a field line, so the upstream point is at higher potential.

Surfaces on which $V$ is constant are called equipotentials. Since moving along one costs no energy, $\vec E$ must be perpendicular to them everywhere. Around a point charge they are spheres; near a metal surface they hug the metal.

Spheres and balls

A uniformly charged ball of radius $R$ and total charge $Q$ has, from Gauss's law, $E = kQ/r^2$ outside and $E = kQr/R^3$ inside. Outside, the potential is therefore identical to a point charge's: $V = kQ/r$. A sphere with $8.0\ \text{nC}$ and $R = 6.0\ \text{cm}$, measured at $r = 24\ \text{cm}$, gives $V = (8.99\times10^{9})(8\times10^{-9})/0.24 = 71.92/0.24 = 300\ \text{V}$: the radius never enters. A Van de Graaff dome of radius $25\ \text{cm}$ holding $1.5\ \mu\text{C}$ sits at $(8.99\times10^{9})(1.5\times10^{-6})/0.25 = 53\,940\ \text{V}$.

To go inside, integrate the interior field from the surface inwards:

The centre of a uniformly charged ball sits $kQ/2R$ above its surface. For $R = 20\ \text{cm}$ and $Q = 50\ \text{nC}$: $(8.99\times10^{9})(50\times10^{-9})/(2\times0.20) = 449.5/0.4 = 1124\ \text{V}$.

A conductor is different: charge lives on its surface, the field inside vanishes, and so the whole conductor is one equipotential at $V = kQ/R$.

Example 3 (why sharp points spark). Two metal spheres, radii $2.0\ \text{cm}$ and $9.0\ \text{cm}$, sit far apart and are joined by a thin wire. What is the ratio of the surface fields?

The wire forces them to the same potential: that is the whole content of the problem. So $kQ_1/r_1 = kQ_2/r_2$, which gives $Q_1/Q_2 = r_1/r_2$: the big sphere hoards most of the charge. But the surface field is $E = kQ/r^2$, so

The smaller sphere has the stronger field at its surface, by the ratio of the radii. That is why air breaks down first at a sharp point on a charged object, and why lightning rods are pointed.

Energy of an assembly

Bring charges in one at a time from infinity. The first one is free: there is no field yet. The second arrives in the first one's potential, costing $q_2 V_1 = kq_1q_2/r_{12}$. The third pays for both charges already placed. Add it up and every pair has been counted once:

For two charges $2.0\ \text{nC}$ and $5.0\ \text{nC}$ at $15\ \text{cm}$: $U = (8.99\times10^{9})(2\times10^{-9})(5\times10^{-9})/0.15 = 8.99\times10^{-8}/0.15 = 5.99\times10^{-7}\ \text{J}$. For three $4.0\ \text{nC}$ charges on an equilateral triangle of side $0.30\ \text{m}$ there are three identical pairs: $U = 3(8.99\times10^{9})(16\times10^{-18})/0.30 = 3(4.79\times10^{-7}) = 1.44\times10^{-6}\ \text{J}$, and that is exactly the work you must do to assemble them.

Counting pairs is the whole skill. On a cube of edge $a$ with eight identical charges there are $\binom82 = 28$ pairs: 12 along edges at distance $a$, 12 face diagonals at $a\sqrt2$, and 4 body diagonals at $a\sqrt3$. So

Example 4 (a continuous body). Build up a nonconducting ball of radius $R = 10\ \text{cm}$ to a final charge $Q = 2.0\ \mu\text{C}$, uniformly through its volume. How much work does it take?

You cannot count pairs here, so build the ball in shells. Suppose you have got as far as radius $r$, carrying $q = Q r^3/R^3$. Its surface is at potential $kq/r$. The next shell of thickness $dr$ carries $dq = \rho\,4\pi r^2\,dr = 3Q r^2 dr/R^3$, and bringing it in from infinity costs $dW = (kq/r)\,dq$:

Integrate from $0$ to $R$, where $\int_0^R r^4 dr = R^5/5$:

Now the numbers. $Q^2 = 4.0\times10^{-12}\ \text{C}^2$, so $3kQ^2 = 3(8.99\times10^{9})(4.0\times10^{-12}) = 0.10788$. Divide by $5R = 0.50\ \text{m}$: $W = 0.216\ \text{J}$. That is a lot of energy for two microcoulombs, which is why charged objects in the real world hold so little charge.

Putting it together: energy conservation problems

The best problems combine the two halves. Find $V$ where the particle starts and where it ends, turn the difference into energy, and let kinetic energy take up the slack.

Example 5. A ring of radius $12\ \text{cm}$ carries $+40\ \text{nC}$. An electron is released from rest very far away on the axis. How fast is it going at the centre?

Far away, $V = 0$ and $U = 0$ and $K = 0$. At the centre, $z = 0$ so $V = kQ/R$, and the electron's energy there is $U = (-e)(kQ/R)$: negative, since the electron is attracted. Energy conservation, $K_f + U_f = 0$, gives

Numerically: $2ek = 2(1.6\times10^{-19})(8.99\times10^{9}) = 2.877\times10^{-9}$; times $Q = 40\times10^{-9}$ gives $1.151\times10^{-16}$. The denominator is $mR = (9.11\times10^{-31})(0.12) = 1.093\times10^{-31}$. The quotient is $1.053\times10^{15}$, whose square root is $3.24\times10^{7}\ \text{m/s}$.

The same ring answers a harder question. Displace the electron a small distance $z \ll R$ from the centre and let go. Its energy is $U(z) = -ekQ/\sqrt{z^2+R^2}$, and for small $z$, $(z^2+R^2)^{-1/2} \approx R^{-1}(1 - z^2/2R^2)$, so

A $U$ quadratic in $z$ is a spring with $\tfrac12 k_{\text{eff}} z^2$, so $k_{\text{eff}} = ekQ/R^3$ and $\omega = \sqrt{ekQ/mR^3}$. For $R = 8.0\ \text{cm}$ and $Q = 25\ \text{nC}$: $ekQ = (1.6\times10^{-19})(8.99\times10^{9})(25\times10^{-9}) = 3.596\times10^{-17}$, and $mR^3 = (9.11\times10^{-31})(5.12\times10^{-4}) = 4.664\times10^{-34}$. The ratio is $7.71\times10^{16}$, $\omega = 2.78\times10^{8}\ \text{rad/s}$, and $f = \omega/2\pi = 4.4\times10^{7}\ \text{Hz}$.

One more pattern worth naming: minimum launch speed. Fire a proton from far away at a pair of fixed positive charges. It slows down as it climbs. To just reach a target point, all its kinetic energy must be spent there, so

For two $+5.0\ \text{nC}$ charges $20\ \text{cm}$ apart and the target at the midpoint, each charge is $0.10\ \text{m}$ away, so $V = 2(8.99\times10^{9})(5\times10^{-9})/0.10 = 899\ \text{V}$. Then $\tfrac12 m v^2 = (1.6\times10^{-19})(899) = 1.438\times10^{-16}\ \text{J}$, and $v = \sqrt{2(1.438\times10^{-16})/1.67\times10^{-27}} = \sqrt{1.72\times10^{11}} = 4.2\times10^{5}\ \text{m/s}$. The perpendicular bisector in the problem statement is a red herring for the energy: the path does not matter, only the endpoint potentials do.

How to know this is the tool

Reach for potential when you see any of these.

• The question asks for a speed, a work, or an energy and gives you charges and distances. Energy methods skip the trajectory entirely. If a problem describes a complicated path and then asks only about the endpoints, it is telling you the path is irrelevant.
• "Released from rest", "fired from far away", "just barely reaches". These are energy-conservation signals. "Just barely reaches" means $v = 0$ there.
• You need a field but the geometry has no symmetry Gauss's law can use. Compute $V$ by adding scalars, then differentiate. The ring's axial field falls out of $V = kQ/\sqrt{z^2+R^2}$ in one derivative.
• Several charges and a single point. Adding $kq_i/r_i$ with signs is far faster than adding vectors, and a question that asks for $V$ at the centre of something symmetric is usually trivial.
• Conductors joined by a wire. Same potential is the equation you need.
• A formula for $V(x)$ is handed to you. Differentiate and negate; nothing else is being asked.

And the reverse: if the question asks for a force, a direction, or a torque, potential alone is not enough. You need $\vec E$, which means either Gauss's law or a derivative of $V$.

Next you will take this to metal: what happens when charge is free to move, why every conductor is one equipotential, and how the ratio of charge to potential defines capacitance.