Question

The electric field intensity at a point P due to point charge q kept at a point Q is 24 N $C^{-1}$ and the electric potential at point P due to same charge is 12 J $C^{-1}$ .The order of magnitude of charge q is

• A. ${10}^{-6}$ C
• B. ${10}^{-7}$ C
• C. ${10}^{-10}$ C
• D. ${10}^{-9}$ C

Answer 1

The correct option is D ${10}^{-9}$ C

Given,

$E=24N/C$

$V=12J/C$

$q=?$

The electric field at point $P$

$E=\frac{1}{4\pi {\epsilon }_0}\frac{q}{r}$. . . . . . . . .(1)

Electric voltage at point $P$ is

$V=\frac{1}{4\pi {\epsilon }_0}\frac{q}{r}$. . . . . .(2)

Divide equation (2) by (1), we get

$\frac{V}{E}=\frac{q/4\pi {\epsilon }_{0}r}{q/4\pi {\epsilon }_0{r}^2}=r$

$r=\frac{V}{E}$

$r=\frac{12}{24}=0.5m$

From equation (2),

$q=4\pi {\epsilon }_{0}r.V$

$q=\frac{1}{9\times {10}^9}\times 0.5\times 12$

$q=0.66\times {10}^{-9}\approx {10}^{-9}C$

The correct option is D.