Question

An electrostatic field line leaves at an angle $\alpha$ from point charge $q_1$ and connects with point charge $-q_2$ a tan angle $\beta$($q_1$ and $q_2$ are positive) see figure below. If $q_2=\frac{3}{2}{q}_1$ and $\alpha =30^{\circ },$ then

• A. $0°<\beta <30°$
• B. $\beta =30^{\circ }$
• C. $30^{\circ }<\beta \le 60^{\circ }$
• D. $60^{\circ }<\beta \le 90^{\circ }$

Answer 1

Answer: (A)

Consider another symmetric field line below line joining centres of charges $q_1$ and $q_2$

In given situation, flux (field lines) leaving charge $q_1$ at a solid angle
$2\alpha =$ flux terminating over charge $q_2$ at a solid angle$2\beta$.
Clearly, in given situation

We can say that, flux leaving charge $q^1$ through a cone of semi-vertical angle $\alpha =$ flux terminating on charge $q_2$ through a cone of semi-vertical angle $\beta$..

To calculate flux, we first find flux through an elemental ring of base of cone and then we integrate to get total flux.

Area of elemental ring,
$dA=2\pi rds=2\pi R\sin \alpha •Rd\alpha$
Flux through elemental ring $(E||dA)$
$d\varphi =\frac{kQ}{R^2}\cdot 2\pi R^2\sin \alpha d\alpha$
Total flux through base of cone
$\varphi =\underset{o}{\overset{\alpha }{\int }}\frac{kQ}{R^2}\cdot 2\pi R^2\sin \alpha d\alpha$
$=kQ2\pi \underset{o}{\overset{\alpha }{\int }}\sin \alpha d\alpha$
$\varphi =\frac{Q}{2{\epsilon }_o}.\left(1-cos\alpha \right)$
So, equating flux of both cones, we get
$\frac{q_1}{2{\epsilon }_o}\left(1-\cos \alpha \right)=\frac{q_2}{2{\epsilon }_o}\left(1-\cos \beta \right)K$
$\Rightarrow q_1\left(1-\cos \alpha \right)=\frac{3}{2}{q}_1\left(1-\cos \beta \right)$
Substituting $\alpha =30°$ in above equation, we get
$\Rightarrow \frac{2}{3}\left(1-\cos 30^{\circ }\right)=1-\cos \beta$
$\Rightarrow \frac{2}{3}\left(1-\sqrt{\frac{3}{2}}\right)=1-\cos \beta$
$\cos \beta =1-\frac{2}{3}\left(1-\sqrt{\frac{3}{2}}\right)$
$\Rightarrow \cos \beta \approx 0.9$
So, angle $\beta$ is not more than $30°.$