Question

An infinite number of electric charges each equal to $5\times {10}^{-9}C$ are placed along X – axis at x = 1 cm, x = 3 cm, x = 9 cm, x =27 cm ………. And so on. If the consecutive charges have opposite sign,
Then the electric field in $\frac{N}{C}$ 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. $33\times {10}^4$
• D. $41\times {10}^4$

Answer 1

The correct option is D $41\times {10}^4$

$E=E_1+E_2+E_3+---$
$=\frac{1}{4\pi {ϵ}_0}[\frac{q_1}{r_1^2}-\frac{q_2}{r_2^2}+\frac{q_3}{r_3^2}-\frac{q_4}{r_4^2}+---]$
$=5\times {10}^{-9}\times 9\times {10}^9[\frac{1}{1}-\frac{1}{3^2}+\frac{1}{9^2}-\frac{1}{{27}^2}+---]{10}^4$
$=45\times {10}^4[1-\frac{1}{3^2}+\frac{1}{3^4}-\frac{1}{3^6}+---]$
$=45\times {10}^4\left[\frac{1}{1+\frac{1}{9}}\right]=45\times {10}^4\times \frac{9}{10}$
$41\times {10}^{4}N/C$