An infinite number of electric charges, each | vedclass

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Question

(C) The electric field intensity due to a point charge is given by $E = \frac{1}{4 \pi \varepsilon_0} \cdot \frac{Q}{r^2}$.Since the consecutive charg

Vedclass · Accepted Answer

$36 	imes 10^4$

Vedclass · Answer

(C) The electric field intensity due to a point charge is given by $E = \frac{1}{4 \pi \varepsilon_0} \cdot \frac{Q}{r^2}$.Since the consecutive charges are of opposite signs, the net electric field at $x = 0$ is the sum of fields due to individual charges:$E = \frac{1}{4 \pi \varepsilon_0} \left[ \frac{Q}{r_1^2} - \frac{Q}{r_2^2} + \frac{Q}{r_3^2} - \frac{Q}{r_4^2} + \dots \infty ight]$$E = \frac{Q}{4 \pi \varepsilon_0} \left[ \frac{1}{(1 	imes 10^{-2})^2} - \frac{1}{(2 	imes 10^{-2})^2} + \frac{1}{(4 	imes 10^{-2})^2} - \frac{1}{(8 	imes 10^{-2})^2} + \dots \infty ight]$$E = (9 	imes 10^9) 	imes (5 	imes 10^{-9}) 	imes 10^4 \left[ \frac{1}{1^2} - \frac{1}{2^2} + \frac{1}{4^2} - \frac{1}{8^2} + \dots \infty ight]$$E = 45 	imes 10^4 \left[ 1 - \frac{1}{4} + \frac{1}{16} - \frac{1}{64} + \dots \infty ight]$This is an infinite geometric progression with first term $a = 1$ and common ratio $r = -\frac{1}{4}$.The sum $S_{\infty} = \frac{a}{1 - r} = \frac{1}{1 - (-1/4)} = \frac{1}{5/4} = \frac{4}{5}$.Therefore, $E = 45 	imes 10^4 	imes \frac{4}{5} = 36 	imes 10^4 	ext{ N/C}$.

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