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(C) The electric field $E$ at the origin due to a point charge $q$ at distance $r$ is given by $E = k \frac{q}{r^2}$,where $k = 9 	imes 10^9 \, N \cd

Vedclass · Accepted Answer

$36 	imes 10^4$

Vedclass · Answer

(C) The electric field $E$ at the origin due to a point charge $q$ at distance $r$ is given by $E = k \frac{q}{r^2}$,where $k = 9 	imes 10^9 \, N \cdot m^2/C^2$.Since the charges alternate in sign,the field at $x = 0$ is:$E = k \cdot q \left[ \frac{1}{(1 	imes 10^{-2})^2} - \frac{1}{(2 	imes 10^{-2})^2} + \frac{1}{(4 	imes 10^{-2})^2} - \frac{1}{(8 	imes 10^{-2})^2} + \dots ight]$$E = (9 	imes 10^9) 	imes (5 	imes 10^{-9}) 	imes 10^4 \left[ 1 - \frac{1}{4} + \frac{1}{16} - \frac{1}{64} + \dots ight]$$E = 45 	imes 10^4 \left[ 1 - \frac{1}{4} + \frac{1}{16} - \frac{1}{64} + \dots ight]$This is a geometric series with first term $a = 1$ and common ratio $r = -1/4$.The sum $S = \frac{a}{1 - r} = \frac{1}{1 - (-1/4)} = \frac{1}{5/4} = \frac{4}{5}$.$E = 45 	imes 10^4 	imes \frac{4}{5} = 9 	imes 10^4 	imes 4 = 36 	imes 10^4 \, N/C$.

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