Question

Find the magnitude of a charge which produces an electric field of strenght is 18 x ${10}^{3}N{C}^{-1}$ at a distance of 5 m in air.

• A. 50 $\mu$c
• B. 100 $\mu$c
• C. 25 $\mu$c
• D. 250 $\mu$c

Answer 1

The correct option is A 50 $\mu$c

As $E=\frac{1}{4\pi {ϵ}_0}\frac{q}{r^2}$

Given $E=18\times {10}^{3}N{C}^{-1}$ ; r=5m ; $\frac{1}{4\pi {ϵ}_0}=9\times {10}^9$

Putting up values

$18\times {10}^3=9\times {10}^9\frac{q}{25}\Rightarrow q=50\times {10}^{-6}C$

$q=50\mu C$