An electrostatic field line leaves at an angl | vedclass

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(A) The electric flux $\phi$ originating from a point charge $q$ within a cone of semi-vertical angle $	heta$ is given by $\phi = \frac{q}{2\varepsil

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$0^{\circ} < \beta < 30^{\circ}$

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(A) The electric flux $\phi$ originating from a point charge $q$ within a cone of semi-vertical angle $	heta$ is given by $\phi = \frac{q}{2\varepsilon_{0}}(1 - \cos 	heta)$.Since the field lines originating from $q_{1}$ within the cone of angle $\alpha$ terminate on $q_{2}$ within the cone of angle $\beta$,the flux must be equal:$\frac{q_{1}}{2\varepsilon_{0}}(1 - \cos \alpha) = \frac{q_{2}}{2\varepsilon_{0}}(1 - \cos \beta)$Given $q_{2} = \frac{3}{2} q_{1}$,we have:$q_{1}(1 - \cos \alpha) = \frac{3}{2} q_{1}(1 - \cos \beta)$$1 - \cos 30^{\circ} = \frac{3}{2}(1 - \cos \beta)$$1 - \frac{\sqrt{3}}{2} = \frac{3}{2}(1 - \cos \beta)$$1 - 0.866 = 1.5(1 - \cos \beta)$$0.134 = 1.5(1 - \cos \beta)$$1 - \cos \beta = \frac{0.134}{1.5} \approx 0.0893$$\cos \beta = 1 - 0.0893 = 0.9107$Since $\cos 30^{\circ} \approx 0.866$,and $\cos \beta \approx 0.9107$,we find that $\beta Thus,$0^{\circ} < \beta < 30^{\circ}$.

An electrostatic field line leaves at an angle $\alpha$ from a point charge $q_{1}$ and connects with a point charge $-q_{2}$ at an angle $\beta$ ($q_{1}$ and $q_{2}$ are positive). See the figure below. If $q_{2} = \frac{3}{2} q_{1}$ and $\alpha = 30^{\circ}$,then:

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