Question

An infinite number of electric charges each equal to 5 nano-coulomb (magnitude) are placed along X-axis at x = 1 cm, x = 2 cm, x = 4 cm, x = 8 cm ......... and so on. In the setup if the consecutive charges have opposite sign, then the electric field in Newton/Coulomb at x = 0 is $(\frac{1}{4\pi {ϵ}_0}=9\times {10}^{9}N-m^2/c^2)$

• A. $12\times {10}^4$
• B. $24\times {10}^4$
• C. $36\times {10}^4$
• D. $48\times {10}^4$

Answer 1

The correct option is C $36\times {10}^4$

$E=\frac{1}{4\pi {ϵ}_0}[\frac{5\times {10}^{-9}}{(1\times {10}^{-2}{)}^2}-\frac{5\times {10}^{-9}}{(2\times {10}^{-2}{)}^2}+\frac{5\times {10}^{-9}}{(4\times {10}^{-2}{)}^2}-\frac{(5\times {10}^{-9})}{(8\times {10}^{-2}{)}^2}+\dots \dots ]$
$\Rightarrow E=\frac{9\times {10}^9\times 5\times {10}^{-9}}{{10}^{-4}}[1-\frac{1}{(2{)}^2}+\frac{1}{(4{)}^2}-\frac{1}{(8{)}^2}+\dots \dots ]$
$\Rightarrow E=45\times {10}^4[1+\frac{1}{(4{)}^2}+\frac{1}{(16{)}^2}+\dots \dots ]-45\times {10}^4[1+\frac{1}{(2{)}^2}+\frac{1}{(8{)}^2}+\frac{1}{(32{)}^2}\dots \dots ]$
$\Rightarrow E=48\times {10}^4-12\times {10}^4=36\times {10}^{4}N/C$