Question

What will be the electric field, if in the above set up, the consecutive charges have opposite charges?

• A. $\frac{q}{4\pi {ϵ}_0}$
• B. $\frac{q}{3\pi {ϵ}_0}$
• C. $\frac{q}{2\pi {ϵ}_0}$
• D. $\frac{q}{5\pi {ϵ}_0}$

Answer 1

The correct option is D $\frac{q}{5\pi {ϵ}_0}$

$E=\frac{q}{4\pi {ϵ}_o{x}^2}$

$E_{net}=\frac{q}{4\pi {ϵ}_o}(\frac{1}{1^2}-\frac{1}{2^2}+\frac{1}{4^2}-...)$ since consecutive charge of opposite nature

Now, $\frac{1}{1^2}-\frac{1}{2^2}+\frac{1}{4^2}-\frac{1}{8^2}+...$

$=\frac{1}{1}-\frac{1}{4}+\frac{1}{16}-\frac{1}{64}+...$

This is an infinite geometric series of common ratio $R=\frac{-1}{4}>-1$

$=\frac{1}{1-\frac{-1}{4}}=\frac{4}{5}$

Hence,$E_{net}=\frac{q}{4\pi {ϵ}_o}\frac{4}{5}$

$⟹E_{net}=\frac{q}{5\pi {ϵ}_o}$

Answer-(D)