Worked examples · Sheet 1

Potentials of systems of point charges

Theory used: Potential of a point charge Potential of a system of charges Electric potential Equipotential surfaces Field as the gradient of the potential

Go to: Exercise 1 Exercise 2 Exercise 3

Exercise 1

A point charge $q_{1} = + 4.0 nC$ is at rest at the origin and a point charge $q_{2} = - 2.0 nC$ is at rest at the point $(0.30 m, 0)$ . Compute the potential at the midpoint $M$ of the segment joining the charges and at the point $P = (0, 0.40 m)$ , with zero potential at infinite distance.

V (P) = \frac{k q _{1}}{r _{1}} + \frac{k q _{2}}{r _{2}}

r_{1 M} = r_{2 M} = 0.15 m, r_{1 P} = 0.40 m, r_{2 P} = 0.3 0^{2} + 0.4 0^{2} m = 0.50 m

V_{M} V_{P} = \frac{k ( q _{1} + q _{2} )}{0.15 m} = \frac{8.99 \cdot 1 0 ^{9} \cdot 2.0 \cdot 1 0 ^{- 9}}{0.15} V \approx 1.2 \cdot 1 0^{2} V = 8.99 \cdot 1 0^{9} (\frac{4.0 \cdot 1 0 ^{- 9}}{0.40} - \frac{2.0 \cdot 1 0 ^{- 9}}{0.50}) V = 8.99 \cdot 1 0^{9} \cdot 6.0 \cdot 1 0^{- 9} V \approx 54 V

The charges are pointlike and the reference is at infinity, so each produces the potential $\frac{k q}{r}$ , and by the potential of a system of charges the total potential is the sum of the two contributions, with their signs. At the midpoint the distances are equal, therefore the two terms add over a common denominator and only the total charge $q_{1} + q_{2} = 2.0 nC$ matters. At $P$ the distance from $q_{2}$ is the hypotenuse of the right triangle with legs $0.30 m$ and $0.40 m$ , that is, $0.50 m$ . Both potentials are positive, because at both points the contribution of the positive charge prevails, since it is larger in absolute value and closer or at the same distance.

Exercise 2

With the same charges as in Exercise 1, determine the points of the $x$ axis where the potential is zero. Then show that in the $x y$ plane the points at zero potential form a circle.

V (x) = k (\frac{q _{1}}{∣ x ∣} + \frac{q _{2}}{∣ x - 0.30∣}) = 0 ⟺ 2 ∣ x - 0.30∣ = ∣ x ∣, x \neq = 0, x \neq = 0.30

for $x < 0$ : $2 (0.30 - x) = - x$ , that is, $x = 0.60$ , which does not satisfy $x < 0$ ;
for $0 < x < 0.30$ : $2 (0.30 - x) = x$ , that is, $x = 0.20 m$ ;
for $x > 0.30$ : $2 (x - 0.30) = x$ , that is, $x = 0.60 m$ .

V (x, y) = 0 ⟺ r_{1} = 2 r_{2} ⟺ x^{2} + y^{2} = 4 [(x - 0.30)^{2} + y^{2}] ⟺ 3 x^{2} - 2.4 x + 0.36 + 3 y^{2} = 0 ⟺ (x - 0.40)^{2} + y^{2} = 0.2 0^{2}

The line at zero potential is the circle with center

(0.40, 0)

and radius

0.20 m

, which cuts the

x

axis at

0.20 m

and

0.60 m

and encloses the negative charge

The potential is zero when the two contributions are opposite. Since $q_{1} = - 2 q_{2}$ , the condition is equivalent to $r_{1} = 2 r_{2}$ , that is, the point must be twice as far from the positive charge as from the negative one. On the $x$ axis the distances are absolute values, so the three intervals separated by the positions of the charges are considered separately. To the left of the origin the solution found does not belong to the interval, and indeed there every point is closer to the positive charge, which is also the larger in absolute value; therefore the potential vanishes only at the points $x = 0.20 m$ and $x = 0.60 m$ .

In the plane the condition $r_{1} = 2 r_{2}$ is squared, since the distances are positive, and after dividing by $3$ and completing the square in $x$ it becomes the equation of a circle with center $(0.40, 0)$ and radius $0.20 m$ . By the symmetry with respect to the $x$ axis, in space the equipotential surface $V = 0$ is the sphere with the same center and the same radius. As a check, the two points of the axis found before are the endpoints of the horizontal diameter.

Exercise 3

Two point charges $q_{1} = q_{2} = + 2.0 nC$ occupy two vertices of an equilateral triangle of side $a = 0.30 m$ . Determine the charge $q_{3}$ to be placed at the third vertex for the potential at the centroid $G$ of the triangle to be zero. With this charge, compute the potential at the midpoint $M$ of the side joining $q_{1}$ and $q_{2}$ and the electric field at $G$ .

V (G) = \frac{k ( q _{1} + q _{2} + q _{3} )}{d _{G}} = 0, d_{G} = \frac{a}{3} ⟹ q_{3} = - (q_{1} + q_{2}) = - 4.0 nC

V (M) = \frac{k q _{1}}{\frac{a}{2}} + \frac{k q _{2}}{\frac{a}{2}} + \frac{k q _{3}}{h}, h = \frac{3}{2} a

h V (M) = 0.866 \cdot 0.30 m \approx 0.260 m = 8.99 \cdot 1 0^{9} (\frac{4.0 \cdot 1 0 ^{- 9}}{0.15} - \frac{4.0 \cdot 1 0 ^{- 9}}{0.260}) V = 8.99 \cdot 1 0^{9} \cdot 1.13 \cdot 1 0^{- 8} V \approx 1.0 \cdot 1 0^{2} V

E (G) = 2 \cdot \frac{k q _{1}}{d _{G}^{2}} cos 6 0^{\circ} + \frac{k ∣ q _{3} ∣}{d _{G}^{2}} = \frac{k ( q _{1} + ∣ q _{3} ∣ )}{d _{G}^{2}} = \frac{3 k ( q _{1} + ∣ q _{3} ∣ )}{a ^{2}}

E (G) = \frac{3 \cdot 8.99 \cdot 1 0 ^{9} \cdot 6.0 \cdot 1 0 ^{- 9}}{0.3 0 ^{2}} V/m \approx 1.8 \cdot 1 0^{3} V/m

The centroid of an equilateral triangle is equidistant from the three vertices, at distance $\frac{a}{3}$ , so by the potential of a system of charges the potential at $G$ is proportional to the total charge, and it vanishes if $q_{3} = - 4.0 nC$ . At $M$ the charges $q_{1}$ and $q_{2}$ are half a side away, while $q_{3}$ is at the distance $h$ of the altitude; the positive contribution prevails and the potential is about $100 V$ .

The field at $G$ is computed as a sum of vectors. The field of each positive charge is directed from that charge toward $G$ and continues beyond it; the two fields make an angle of $6 0^{\circ}$ with the median going from $M$ to the third vertex, therefore the components perpendicular to the median cancel and the two components $\frac{k q _{1}}{d _{G}^{2}} cos 6 0^{\circ}$ directed toward $q_{3}$ remain. The field of $q_{3}$ , which is negative, is directed toward $q_{3}$ , so all the contributions have the same direction. The field at $G$ is therefore nonzero, about $1.8 \cdot 1 0^{3} V/m$ toward the vertex of $q_{3}$ , even though the potential at $G$ is zero. Indeed, by the field as the gradient of the potential, the field at a point depends on how the potential varies around the point.