Looking for classes? Ksquare Career Institute, Bengaluru →
The electric potential at a point in free space due to a charge $Q$ coulomb is $Q \times 10^{11}$ volts. The electric field at that point is -
A
$4\pi\varepsilon_0 Q \times 10^{20}$ volt/m
B
$12\pi\varepsilon_0 Q \times 10^{22}$ volt/m
C
$4\pi\varepsilon_0 Q \times 10^{22}$ volt/m
D
$12\pi\varepsilon_0 Q \times 10^{20}$ volt/m
Detailed Solution
For a point charge: $V = \dfrac{Q}{4\pi\varepsilon_0 r}$ and $E = \dfrac{Q}{4\pi\varepsilon_0 r^2}$
Given $V = Q \times 10^{11} = \dfrac{Q}{4\pi\varepsilon_0 r}$
$\Rightarrow r = \dfrac{1}{4\pi\varepsilon_0 \times 10^{11}}$
Dividing the two expressions: $E = \dfrac{V}{r}$
$E = Q \times 10^{11} \times 4\pi\varepsilon_0 \times 10^{11}$
$E = 4\pi\varepsilon_0 Q \times 10^{22}$ volt m$^{-1}$
Given $V = Q \times 10^{11} = \dfrac{Q}{4\pi\varepsilon_0 r}$
$\Rightarrow r = \dfrac{1}{4\pi\varepsilon_0 \times 10^{11}}$
Dividing the two expressions: $E = \dfrac{V}{r}$
$E = Q \times 10^{11} \times 4\pi\varepsilon_0 \times 10^{11}$
$E = 4\pi\varepsilon_0 Q \times 10^{22}$ volt m$^{-1}$