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(A) The electric field $E$ at the origin due to a point charge $q$ at distance $x$ is given by $E = \frac{kq}{x^2}$.Since all charges $q$ are placed a

Vedclass · Accepted Answer

$\frac{4kq}{3}$

Vedclass · Answer

(A) The electric field $E$ at the origin due to a point charge $q$ at distance $x$ is given by $E = \frac{kq}{x^2}$.Since all charges $q$ are placed at positions $x = 1, 2, 4, 8, ... 	ext{ m}$,the total electric field at the origin is:$E_0 = \frac{kq}{1^2} + \frac{kq}{2^2} + \frac{kq}{4^2} + \frac{kq}{8^2} + ...$$E_0 = kq \left[ 1 + \frac{1}{4} + \frac{1}{16} + \frac{1}{64} + ... ight]$This is an infinite geometric series with first term $a = 1$ and common ratio $r = \frac{1}{4}$.The sum of an infinite geometric series is $S_{\infty} = \frac{a}{1 - r}$.$S_{\infty} = \frac{1}{1 - 1/4} = \frac{1}{3/4} = \frac{4}{3}$.Therefore,$E_0 = kq \left( \frac{4}{3} ight) = \frac{4kq}{3}$.

Calculate the electric field at the origin due to an infinite number of charges as shown in the figure.

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