Question

The 3 charges → \(\frac {q}{2}\), -q, \(\frac {q}{2}\) are placed along the x axis at x = 0, x = r, x = 2 respectively. Find the resultant potential at a point A located at a distance y from charge q such that r << y.

(a) V=\(\frac {4qr^2}{4\pi \varepsilon_o y}\)

(b) V=\(\frac {qr^3}{4\pi \varepsilon_o y^3}\)

(c) V=\(\frac {qr^2}{4\pi \varepsilon_o y^3}\)

(d) V=\(\frac {qr^3}{4\pi \varepsilon_o y^2}\)

I got this question during an interview.

Question is taken from Electrostatic Potential due to a System of Charges in division Electrostatic Potential and Capacitance of Physics – Class 12

Answer 1

Correct choice is (c) V=\(\frac {qr^2}{4\pi \varepsilon_o y^3}\)

Easiest explanation: Electric potential at point A can be given by:

V=\(\frac {1}{4\pi \varepsilon_o}[(\frac {(\frac {q}{2})}{y+r}) – \frac {q}{y} + \frac {(\frac {q}{2})}{(y-r)}]\)

V=\(\frac {(qr^2)}{4\pi \varepsilon_o y(y^2 – r^2)}\)

V=\(\frac {qr^2}{4\pi \varepsilon_o y^3}\)                   (Since y >> r)