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(A) The electric field $E$ at $x = 0$ due to charges $Q$ at positions $x_n = 2^n$ (for $n = 0, 1, 2, \dots$) is given by the sum of individual fields:

Vedclass · Accepted Answer

$12 	imes 10^9 Q 	ext{ N/C}, 1.8 	imes 10^4 	ext{ V}$

Vedclass · Answer

(A) The electric field $E$ at $x = 0$ due to charges $Q$ at positions $x_n = 2^n$ (for $n = 0, 1, 2, \dots$) is given by the sum of individual fields:$E = \sum_{n=0}^{\infty} \frac{kQ}{x_n^2} = kQ \left[ \frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{4^2} + \frac{1}{8^2} + \dots ight]$$E = kQ \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 is $S = \frac{a}{1-r} = \frac{1}{1 - 1/4} = \frac{4}{3}$.$E = (9 	imes 10^9) 	imes Q 	imes \frac{4}{3} = 12 	imes 10^9 Q 	ext{ N/C}$.The electric potential $V$ at $x = 0$ is given by the sum of individual potentials:$V = \sum_{n=0}^{\infty} \frac{kQ}{x_n} = kQ \left[ \frac{1}{1} + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \dots ight]$This is a geometric series with $a = 1$ and $r = 1/2$. The sum is $S = \frac{1}{1 - 1/2} = 2$.Assuming $Q = 1 \mu C = 10^{-6} C$ (as implied by the numerical result in the options):$V = (9 	imes 10^9) 	imes (10^{-6}) 	imes 2 = 18 	imes 10^3 = 1.8 	imes 10^4 	ext{ V}$.

Electric charges $Q$ are placed on the $x$-axis at $x = 1, 2, 4, 8, \dots \text{meters}$ respectively. What are the electric field and electric potential at $x = 0$?

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