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An electrostatic field line leaves at an angle $\alpha$ from point charge $q_{1}$ and connects with point charge $-q_{2}$ at an angle $\beta\left(q_{1}\right.$ and $q_{2}$ are positive) see figure below. If $q_{2}=\frac{3}{2} q_{1}$ and $\alpha=30^{\circ}$, then
$0^{\circ} < \beta<30^{\circ}$
$\beta=30^{\circ}$
$30^{\circ} < \beta \leq 60^{\circ}$
$60^{\circ} < \beta \leq 90^{\circ}$
Solution
$(a)$ 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 $\psi _{2}$ 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,
$d A=2 \pi r d s=2 \pi R \sin \alpha \cdot R d \alpha$
Flux through elemental ring $( E \| d A )$
$d \phi=\frac{k Q}{R^{2}} \cdot 2 \pi R^{2} \sin \alpha d \alpha$
Total flux through base of cone
$\phi=\int\limits_{0}^{\alpha} \frac{k Q}{R^{2}} \cdot 2 \pi R^{2} \sin \alpha d \alpha$
$=k Q 2 \pi \int\limits_{0}^{\alpha} \sin \alpha d \alpha$
$\phi=\frac{Q}{2 \varepsilon_{0}} \cdot \cdot(1-\cos \alpha)$
So, equating flux of both cones, we get
$\frac{q_{1}}{2 \varepsilon_{0}}(1-\cos \alpha)=\frac{q_{2}}{2 \varepsilon_{0}}(1-\cos \beta)$
$\Rightarrow \quad q_{1}(1-\cos \alpha) =\frac{3}{2} q_{1}(1-\cos \beta)$
Substituting $\alpha=30^{\circ}$ in above equation, we get
$\Rightarrow \frac{2}{3}\left(1-\cos 30^{\circ}\right) =1-\cos \beta$
$\Rightarrow \frac{2}{3}\left(1-\frac{\sqrt{3}}{2}\right) =1-\cos \beta$
$\Rightarrow \cos \beta =1-\frac{2}{3}\left(1-\frac{\sqrt{3}}{2}\right)$
$\Rightarrow \cos \beta \approx 0.9$
So, angle $\beta$ is not more than $30^{\circ}$.
