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Question

Physics Question on Electric Field

The magnitude of point charge due to which the electric field 30 cm away has the magnitude ,2 NC−12 \, NC^{-1} will be

A

2×10−11 C2 \times 10^{-11} \, C

B

3×10−11 C3 \times 10^{-11} \, C

C

5×10−11 C5 \times 10^{-11} \, C

D

9×10−11 C9 \times 10^{-11} \, C

Answer

2×10−11 C2 \times 10^{-11} \, C

Explanation

Solution

E=kqr2E = \frac{kq}{r^{2}}
q=Er2k=2×(0.3)29×109=2×9×10−2×10−99q = \frac{Er^{2}}{k} = \frac{2 \times\left(0.3\right)^{2}}{9\times10^{9}} = \frac{2\times9 \times10^{-2} \times10^{-9}}{9}
∴q=2×10−11 C\therefore q = 2 \times10^{-11}\, C