Question

The magnitude of point charge due to which the electric field 30 cm away has the magnitude ,$2 \, NC^{-1}$ will be

• A. $ 2 \times 10^{-11} \, C $
• B. $ 3 \times 10^{-11} \, C $
• C. $ 5 \times 10^{-11} \, C $
• D. $ 9 \times 10^{-11} \, C $

Answer 1

Answer: (A)

$E = \frac{kq}{r^{2}} $
$ q = \frac{Er^{2}}{k} = \frac{2 \times\left(0.3\right)^{2}}{9\times10^{9}} = \frac{2\times9 \times10^{-2} \times10^{-9}}{9}$
$ \therefore q = 2 \times10^{-11}\, C$