Exercises
Two charges 5 × 10⁻⁸ C and –3 × 10⁻⁸ C are located 16 cm apart. At what point(s) on the line joining the two charges is the electric potential zero? Take the potential at infinity to be zero.
Given: q1 = +5 × 10⁻⁸ C, q2 = –3 × 10⁻⁸ C, separation d = 16 cm = 0.16 m.
Take q1 at the origin and q2 at x = 0.16 m on the x-axis.
Case 1 — Point P between the charges at distance x from q1:
V = kq1/x + kq2/(d – x) = 0
⇒ 5 × 10⁻⁸ / x = 3 × 10⁻⁸ / (0.16 – x)
⇒ 5(0.16 – x) = 3x ⇒ 0.80 = 8x ⇒ x = 0.10 m = 10 cm
Case 2 — Point P outside the charges, beyond q2, at distance x from q1 (x > 0.16 m):
V = kq1/x + kq2/(x – d) = 0
⇒ 5 × 10⁻⁸ / x = 3 × 10⁻⁸ / (x – 0.16)
⇒ 5(x – 0.16) = 3x ⇒ 2x = 0.80 ⇒ x = 0.40 m = 40 cm
(There is no solution on the side beyond q1 because both terms would have the same sign and cannot cancel.)
Answer: Electric potential is zero at two points: 10 cm from the +5 × 10⁻⁸ C charge (between the charges), and 40 cm from the +5 × 10⁻⁸ C charge (on the side of the negative charge, beyond it).