Question

An electrostatic field line leaves at angle $\alpha$ from point charge $+q_1$, and connects with point charge $–q_2$ at angle $\beta$. Find the ratio $\frac{q_1}{q_2}$ (Given : $\alpha ={60}^{\circ },\beta ={90}^{\circ }$).

• A. $\frac{1}{2}$
• B. $\frac{2}{1}$
• C. $\frac{3}{4}$
• D. None of these

Answer 1

The correct option is B $\frac{2}{1}$

The solid angle substituted by the charge $q_1$ is
${\Omega }_1=2\pi (1-\cos {60}^{\circ })=\pi$
And the solid angle substituted by the charge $-q_2$ is
${\Omega }_2=2\pi (1-\cos {90}^{\circ })=2\pi$
Flux from $q_1=\frac{q_1}{4\pi {ϵ}_0}\times {\Omega }_1=\frac{q_1}{4\pi {ϵ}_0}\times \pi =\frac{q_1}{4{ϵ}_0}$
Flux from $q_2=\frac{q_2}{4\pi {ϵ}_0}\times {\Omega }_2=\frac{q_2}{4\pi {ϵ}_0}\times 2\pi =\frac{q_2}{2{ϵ}_0}$

Outward flux = inward flux
$\frac{q_1}{4{ϵ}_0}=\frac{q_2}{2{ϵ}_0}$
$\frac{q_1}{q_2}=\frac{2}{1}$