A charge + q is fixed at each of the points x = x0,x=3x0,x=5x0...∞ on the x-axis and a charge - q is fixed at each of the points x = 2x0,x=4x0,x=6x0...,∞ Here, x0 is a positive constant. Take the electric potential at a point due to a charge Q at a distance r from it to be Q/4 π ε0 r. Then the potential at the origin due to the above system of charges is

Question

A charge + q is fixed at each of the points x = x0,x=3x0,x=5x0...∞ on the x-axis and a charge - q is fixed at each of the points x = 2x0,x=4x0,x=6x0...,∞ Here, x0 is a positive constant. Take the electric potential at a point due to a charge Q at a distance r from it to be Q/4 π ε0 r. Then the potential at the origin due to the above system of charges is (A) zero (B) (q/8 π ε0x0 ln 2) (C) infinite (D) (q ln 2/4 πε0x0)

Tardigrade Admin · Accepted Answer

Potential at origin will be given by V=(q/4 π ε0) [(1/x0)-(1/2x0)+(1/3x0)-(1/4x0)+....] =(q/4 π ε0).(1/x0)[1-(1/2)+(1/3)-(1/4)+....]=(q/4 π ε0 x0) ln (2)

zero

$\frac{q}{8 π ε _{0} x _{0} l n 2}$

infinite

$\frac{q l n 2}{4 π ε _{0} x _{0}}$

Potential at origin will be given by
$V = \frac{q}{4 π ε _{0}} [\frac{1}{x _{0}} - \frac{1}{2 x _{0}} + \frac{1}{3 x _{0}} - \frac{1}{4 x _{0}} + ....]$
$= \frac{q}{4 π ε _{0}} . \frac{1}{x _{0}} [1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + ....] = \frac{q}{4 π ε _{0} x _{0}} ln (2)$

zero

8πε0​x0​ln2q​

infinite

4πε0​x0​qln2​

Potential at origin will be given by V=4πε0​q​[x0​1​−2x0​1​+3x0​1​−4x0​1​+....] =4πε0​q​.x0​1​[1−21​+31​−41​+....]=4πε0​x0​q​ln(2)

$\frac{q}{8 π ε _{0} x _{0} l n 2}$

$\frac{q l n 2}{4 π ε _{0} x _{0}}$

Potential at origin will be given by
$V = \frac{q}{4 π ε _{0}} [\frac{1}{x _{0}} - \frac{1}{2 x _{0}} + \frac{1}{3 x _{0}} - \frac{1}{4 x _{0}} + ....]$
$= \frac{q}{4 π ε _{0}} . \frac{1}{x _{0}} [1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + ....] = \frac{q}{4 π ε _{0} x _{0}} ln (2)$