Potential of point charges, potential energy and finding E from V

Last week you saw potential in a uniform field. This week you go down to the source: the potential of a point charge. The nice part of this topic is that, where field problems make you deal with vector components, here you just add signed numbers. Three patterns appear on the exam: the potential made by several charges at a point, the potential energy of a system of charges (or the work needed to build it), and deriving the field from a given function $V(x,y,z)$.

Golden rules:

1. $V = kq/r$, with the sign of the charge. The potential of a negative charge is negative.
2. Potentials add as scalars. No angles, no components, no $\cos\theta$.
3. Pair energy is $U = kq_1q_2/r_{12}$. In a system each pair is counted once.
4. $\vec E = -\nabla V$. The field points in the direction in which the potential falls fastest.
5. The formula has $r$, not $r^2$. The $r^2$ is in the formula for the field.

Potential of a point charge

The field of a point charge $q$ is radial: $E = kq/r^2$. Integrating it from infinity to the point $r$ and choosing $V(\infty) = 0$:

$V(r) = -\int_\infty^r \frac{kq}{r'^2}\,dr' = \frac{kq}{r} = \frac{1}{4\pi\varepsilon_0}\frac{q}{r}$

Constants: $k = 8.99\times10^{9}\ \text{N·m}^2/\text{C}^2$, $\varepsilon_0 = 8.85\times10^{-12}\ \text{C}^2/(\text{N·m}^2)$.

The potential falls as $1/r$, the field as $1/r^2$. If you know both at the same point, there is the shortcut $E = V/r$ for a single charge.

Superposition

With more than one charge, compute the potential of each one separately and add them with their signs:

$V_P = k\sum_i \frac{q_i}{r_i}$

$r_i$ is the distance from the $i$th charge to the point P, always positive. The sign is carried by the charge.

PITFALL · Adding potentials like vectors. Four equal charges at the corners of a square make the field zero at the center (symmetry), but not the potential. Four positive contributions add: $V = 4kq/r$. Potential has no direction; cancellation happens only between charges of opposite sign.

Potential energy of charge systems

With $q_1$ fixed, the work an external agent does to bring $q_2$ from infinity to a distance $r_{12}$ is $W = q_2 V_1 = k q_1 q_2 / r_{12}$. This work is stored as the potential energy of the pair:

$U = \frac{k\,q_1 q_2}{r_{12}}$

• Charges of the same sign: $U > 0$. You spent work to bring them together; when released they repel each other and the energy turns into kinetic energy.
• Charges of opposite sign: $U < 0$. The system is bound; work is needed to separate them.

For three or more charges, each pair is counted once:

$U = k\left(\frac{q_1q_2}{r_{12}} + \frac{q_1q_3}{r_{13}} + \frac{q_2q_3}{r_{23}}\right)$

With $n$ charges there are $n(n-1)/2$ pairs: 3 for 3 charges, 6 for 4 charges.

TIP · Adding one charge. The work of bringing a new charge $q$ into an existing system is the potential of the other charges at that point times $q$: $W = qV_P$. New system energy = old energy + $qV_P$. You don’t need to rewrite all the pairs.

Finding the field from the potential

If you reverse the relation $dV = -\vec E\cdot d\vec s$, you get the field as a derivative of the potential:

$E_x = -\frac{\partial V}{\partial x}, \qquad E_y = -\frac{\partial V}{\partial y}, \qquad E_z = -\frac{\partial V}{\partial z}, \qquad \vec E = -\nabla V$

Partial derivative: when you differentiate with respect to $x$, treat $y$ and $z$ as constants. With spherical symmetry one variable is enough: $E_r = -dV/dr$. Check: for $V = kq/r$, $-dV/dr = kq/r^2$, correct.

Consequences:

• The field points in the direction in which the potential decreases (that is why there is a minus sign).
• If $V$ is constant (an equipotential region, the inside of a conductor), $\vec E = 0$.
• The value of $V$ at a point does not determine the field, its slope does.

Worked examples

Example 1. Two charges, one point. $q_1 = +3.0$ nC at the origin, $q_2 = -5.0$ nC at $(0.40, 0)$ m. The point P is $(0, 0.30)$ m. What is the potential at P?

r1 = 0.30 m
r2 = sqrt(0.40² + 0.30²) = 0.50 m
V1 = (8.99e9)(3.0e-9)/0.30 = +89.9 V
V2 = (8.99e9)(-5.0e-9)/0.50 = -89.9 V
VP = V1 + V2 = 0 V

Interesting: the potential is zero but the field is not. If you compute with components you get $E_x = 144$ V/m, $E_y = 192$ V/m, $|\vec E| = 240$ V/m. $V = 0$ does not mean “no field”.

Example 2. A three-charge system. In addition to the same two charges, bring $q_3 = +4.0$ nC to the point P. (a) What work is needed? (b) What is the total energy of the system?

(a) W = q3 · VP = (4.0e-9)(0) = 0 J
(b) U12 = k(3.0e-9)(-5.0e-9)/0.40 = -3.37e-7 J
    U13 = k(3.0e-9)(4.0e-9)/0.30 = +3.60e-7 J
    U23 = k(-5.0e-9)(4.0e-9)/0.50 = -3.60e-7 J
    U = -3.37e-7 J

Check: $U_{13} + U_{23} = q_3V_P = 0$, so the new charge added no energy to the system. The shortcut in the tip agrees.

Example 3. Field from a potential function. $V(x,y) = 3.0x^2y + 2.0y^2$ (V, m). What is the field at $(1.0, 2.0)$ m?

∂V/∂x = 6.0xy        -> Ex = -6.0(1.0)(2.0) = -12 V/m
∂V/∂y = 3.0x² + 4.0y -> Ey = -(3.0 + 8.0)  = -11 V/m
|E| = sqrt(12² + 11²) = 16.3 V/m

If you forget the minus sign, the direction comes out reversed; the magnitude stays the same, but you lose points on a component question.

Example 4. From repulsion to speed. If two positive charges are held fixed and then released, the initial $U$ turns completely into kinetic energy. If the masses are equal, momentum conservation makes the speeds equal and the energy is shared equally between the two particles. If the masses differ, use momentum conservation first ($m_1v_1 = m_2v_2$), then the energy equation.

Example 5. Electric dipole. A $+4.0$ nC charge at $(0.10, 0)$ m and a $-4.0$ nC charge at $(-0.10, 0)$ m. (a) What is the potential on the axis at $x = 0.30$ m? (b) What is the potential at the point $(0, 0.25)$ m on the perpendicular bisector?

(a) r+ = 0.30 - 0.10 = 0.20 m,   r- = 0.30 + 0.10 = 0.40 m
    V = k(4.0e-9)(1/0.20 - 1/0.40) = (35.96)(2.5) = 89.9 V
(b) r+ = r- = sqrt(0.10² + 0.25²) = 0.269 m
    V = kq/r - kq/r = 0 V

The perpendicular bisector plane of a dipole is an equipotential with $V = 0$ throughout. But the field there is not zero: it points from the positive to the negative charge, parallel to the axis. Equipotential surfaces are perpendicular to the field lines; you see this in the dipole picture.

Example 6. Finding the $V = 0$ points. $+4.0$ nC at $x = 0$, $-1.0$ nC at $x = 0.30$ m. Which points on the axis have $V = 0$?

Region between (0 < x < 0.30):  4.0/x = 1.0/(0.30 - x)  ->  x = 0.24 m
Right region (x > 0.30):        4.0/x = 1.0/(x - 0.30)  ->  x = 0.40 m
Left region (x < 0): the larger charge is closer, the total is always positive -> no solution

Rule: the $V = 0$ points are always near the smaller charge, because to make the smaller charge’s contribution equal the larger one’s you have to be closer to it. Don’t confuse this with the point where the field is zero: the field involves squares ($1/r^2$), the potential does not ($1/r$). For the field there may be no solution in the region between at all, because the fields of opposite charges point the same way between them.

Exam strategy

1. Draw the coordinates on paper. Find the distance from each charge to the point with Pythagoras and write it next to the charge.
2. Convert units at the start. nC and cm are the most common source of lost points on the exam.
3. In a potential question, set up a table: charge, distance, $kq/r$. Then add with signs.
4. In an energy question, list the pairs. 3 charges: 12, 13, 23. 4 charges: 12, 13, 14, 23, 24, 34. Don’t count missing or extra pairs.
5. In a work question, check whose work is asked. External agent: $W = q\Delta V$. Field: $W = -q\Delta V$.
6. If $V(x,y)$ is given, differentiate with respect to each variable separately, put in the minus sign, then substitute the point. Don’t substitute the point before differentiating.

Common mistakes