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(B) The electric field $E$ at the origin due to a point charge $q$ at distance $r$ is given by $E = \frac{kq}{r^2}$.Given the charges $q, -q, q, -q, .

Vedclass · Accepted Answer

$\frac{4kq}{5}$

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(B) The electric field $E$ at the origin due to a point charge $q$ at distance $r$ is given by $E = \frac{kq}{r^2}$.Given the charges $q, -q, q, -q, ...$ are placed at distances $x = 1, 2, 4, 8, ...$ meters respectively.The total electric field at the origin is the sum of the fields due to each charge:$E_0 = \frac{kq}{1^2} + \frac{k(-q)}{2^2} + \frac{kq}{4^2} + \frac{k(-q)}{8^2} + ...$$E_0 = kq \left( 1 - \frac{1}{4} + \frac{1}{16} - \frac{1}{64} + ... \right)$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 $S = \frac{a}{1 - r}$.$E_0 = kq \left( \frac{1}{1 - (-1/4)} \right) = kq \left( \frac{1}{1 + 1/4} \right) = kq \left( \frac{1}{5/4} \right) = \frac{4kq}{5}$.

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

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