Question

A charge + q is placed at each of the points x = x₀, x = 3x₀, x = 5x₀,….,and infinitum on the x-axis, and a charge – q is placed at each of the points x = 2x₀, x = 4x₀, x = 6x₀….and infinitum. 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/(4pe₀ r). Then the potential at the origin due to the above system of charge is-

• A. 0
• B. $\frac{q}{8\pi\varepsilon_{0}\log_{e}2}$
• C. 
• D. $\frac{q\log_{e}2}{4\pi\varepsilon_{0}x_{0}}$

Answer 1

Answer: (D)

$\frac{q\log_{e}2}{4\pi\varepsilon_{0}x_{0}}$

Answer 2

V = $\frac { \mathrm { Kq } } { \mathrm { x } _ { 0 } }$ $\left[ 1 - \frac { 1 } { 2 } + \frac { 1 } { 3 } - \frac { 1 } { 4 } + \frac { 1 } { 5 } \ldots .\right]$

log_e (1 + x) = x – + $\frac { x ^ { 3 } } { 3 }$ –……..

x = 1

V = log_e(2)