Infinite charges of magnitude q each are plac | vedclass

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Question

(A) The electric field $E$ at the origin due to a point charge $q$ at distance $r$ is given by $E = \frac{kq}{r^2}$,where $k = \frac{1}{4\pi\epsilon_0

Vedclass · Accepted Answer

$12 	imes 10^9 q \, N/C$

Vedclass · Answer

(A) The electric field $E$ at the origin due to a point charge $q$ at distance $r$ is given by $E = \frac{kq}{r^2}$,where $k = \frac{1}{4\pi\epsilon_0} = 9 	imes 10^9 \, N\cdot m^2/C^2$.Since all charges are positive and located on the positive $X$-axis,the electric field vectors at the origin will all point in the negative $X$-direction.The net electric field $E_{net}$ is the sum of the individual fields:$E_{net} = \sum \frac{kq}{x_i^2} = kq \left( \frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{4^2} + \frac{1}{8^2} + \dots ight)$$E_{net} = kq \left( 1 + \frac{1}{4} + \frac{1}{16} + \frac{1}{64} + \dots ight)$This is an infinite geometric series with the first term $a = 1$ and common ratio $r = \frac{1}{4}$.The sum of an infinite geometric series is $S = \frac{a}{1 - r}$.$S = \frac{1}{1 - 1/4} = \frac{1}{3/4} = \frac{4}{3}$.Therefore,$E_{net} = (9 	imes 10^9) 	imes q 	imes \frac{4}{3} = 12 	imes 10^9 q \, N/C$.

Infinite charges of magnitude $q$ each are placed at $x = 1, 2, 4, 8, ...$ meters on the $X$-axis. The value of the intensity of the electric field at the point $x = 0$ due to these charges will be:

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