An infinite number of electric charges each e | vedclass

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Question

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

Vedclass · Accepted Answer

$36 	imes 10^4$

Vedclass · Answer

(C) The electric field intensity $E$ due to a point charge $Q$ at a distance $r$ is given by $E = \frac{1}{4 \pi \varepsilon_0} \cdot \frac{Q}{r^2}$.Since the charges are placed at $x = 1, 2, 4, 8, \dots 	ext{ cm}$ and have alternating signs, the net electric field at the origin $(x=0)$ is the vector sum of the fields produced by each charge.Let $Q = 5 	imes 10^{-9} 	ext{ C}$. The distances are $r_n = 2^{n-1} 	imes 10^{-2} 	ext{ m}$ for $n = 1, 2, 3, \dots$.The field at $x=0$ is directed towards the negative $X$-axis for positive charges and positive $X$-axis for negative charges. Summing the magnitudes with appropriate signs:$E = \frac{1}{4 \pi \varepsilon_0} \left[ \frac{Q}{(1 	imes 10^{-2})^2} - \frac{Q}{(2 	imes 10^{-2})^2} + \frac{Q}{(4 	imes 10^{-2})^2} - \frac{Q}{(8 	imes 10^{-2})^2} + \dots ight]$$E = \frac{Q}{4 \pi \varepsilon_0 	imes 10^{-4}} \left[ \frac{1}{1^2} - \frac{1}{2^2} + \frac{1}{4^2} - \frac{1}{8^2} + \dots ight]$$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]$The term in the bracket is an infinite geometric series with first term $a = 1$ and common ratio $r = -1/4$.The sum $S = \frac{a}{1-r} = \frac{1}{1 - (-1/4)} = \frac{1}{5/4} = \frac{4}{5}$.$E = 45 	imes 10^4 	imes \frac{4}{5} = 36 	imes 10^4 	ext{ N/C}$.

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