Question

An infinite number of charges each equal to $q$ are placed along the x-axis at $x=1,x=2,x=4,x=8$and so on. Find the potential and electric field at the point $x=0$ due to this set of charges.

• A. $5kq,\frac{4kq}{3}$
• B. $2kq,\frac{4kq}{3}$
• C. 3$kq,\frac{4kq}{3}$
• D. $7kq,\frac{4kq}{4}$

Answer 1

The correct option is B $2kq,\frac{4kq}{3}$

$\text{Step 1: Electric Field at x = 0}$

Electric field at $x=0$ will be sum of electric field due to all charges at $x=1,2,4,8.....$

We know electric field at any point can be given by: $E=\frac{1}{4\pi {\in }_0}\frac{q}{r^2}$

$\therefore E_{x=0}=\frac{1}{4\pi {\in }_0}\frac{q}{1^2}+\frac{1}{4\pi {\in }_0}\frac{q}{2^2}+\frac{1}{4\pi {\in }_0}\frac{q}{4^2}+\frac{1}{4\pi {\in }_0}\frac{q}{8^2}+.....$

$=\frac{1}{4\pi {\in }_0}q[\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{8^2}+.....]$

$E=kq[\frac{1}{1}+\frac{1}{(2{)}^2}+\frac{1}{(2{)}^4}+\frac{1}{(2{)}^6}+.....]$ $....\left(1\right)$ where $k=\frac{1}{4\pi {ϵ}_0}$

Sum of infinite geometric progression $=\frac{a}{1-r}$

$\frac{1}{1}+\frac{1}{(2{)}^2}+\frac{1}{(2{)}^4}+\frac{1}{(2{)}^6}+.....=\frac{1}{1-\frac{1}{4}}=\frac{4}{3}$

Eq. $\left(1\right)$ becomes

$E=\frac{4kq}{3}$

$\text{Step 2: Potenial at x = 0}$

Potential at $x=0$ will be sum of potential due to all charges at $x=1,2,4,8.....$

We know potential at any point can be given by: $V=\frac{1}{4\pi {\in }_0}\frac{q}{r}$

$V{|}_{x=0}=\frac{kq}{1}+\frac{kq}{2}+\frac{kq}{4}+\frac{kq}{8}+.....$

$=kq[1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+.....]$ $....\left(2\right)$

$1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+.....=\frac{1}{1-\frac{1}{2}}=2$

Eq. (2) becomes

$V=2kq$

Hence, Option (B) is correct.