Question

A charge +q is fixed at each of the points x = x₀,

x = 3x₀, x = 5x₀ ,… upto on X-axis and charge –q is fixed on each of the points x = 2x₀, x = 4x₀, x = 6x₀, …upto. Here x₀ is a positive constant. Take the potential at a point due to a charge Q at a distance r from it to be $\frac{Q}{4\pi\varepsilon_{0}r}$. Then the potential at the origin due to above system of charges will be-

• A. zero
• B. $\frac{q}{8\pi\varepsilon_{0}x_{0}\log_{e}2}$
• 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 = $\frac { \mathrm { kq } } { \mathrm { x } _ { 0 } }$ –$\frac { \mathrm { kq } } { 2 \mathrm { x } _ { 0 } } + \frac { \mathrm { kq } } { 3 \mathrm { x } _ { 0 } } - \frac { \mathrm { kq } } { 4 \mathrm { x } _ { 0 } }$+ …….. 

V = $\frac { \mathrm { kq } } { \mathrm { x } _ { 0 } } \left( 1 - \frac { 1 } { 2 } + \frac { 1 } { 3 } - \frac { 1 } { 4 } \ldots \infty \right)$ = $\frac { \mathrm { kq } } { \mathrm { x } _ { 0 } } \left( \log _ { \mathrm { e } } 2 \right)$