Question

Q.48. The electric potential and electric field at any given position due to a point charge are 600 V and 200 N/C respectively. Then magnitude of point charge would be-
(1) 3 $\mu$C
(2) 30 $\mu$C
(3) 0.2 $\mu$C
(4) 0.5 $\mu$C

Answer 1

Dear Student
Given,

Electric field, E = 200 N/C

Electric potential, V = 600 V

Let q be the charge and r be the distance from the charge where the given quantities are observed.

$$\begin{gathered} V=\frac{kq}{r}=600 \\ {V}^2={\left(\frac{kq}{r}\right)}^2=360000 \\ E=\frac{kq}{r^2}=200 \\ \frac{V^2}{E}=\frac{{\left(\frac{kq}{r}\right)}^2}{\frac{kq}{r^2}}=\frac{360000}{200} \\ r=3 \\ V=\frac{kq}{r}=600 \\ kq=1800 \\ \frac{q}{4{\mathrm{\pi \epsilon }}_{\mathrm{o}}}=1800 \\ q=1800\times 4\times 3.14\times 8.85\times {10}^{-12}=0.2\mu C \end{gathered}$$
Regards