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:

Similar Questions

$A$ circular plate sheet of radius $10 \,cm$ is placed in a uniform electric field of $2 \sqrt{3} \times 10^5 \,NC^{-1}$, making an angle of $60^{\circ}$ with the field. Find the electric flux through the sheet.

Which of the following is a correct statement?

Assertion : Electric lines of force never cross each other.
Reason : Electric field at a point superimpose to give one resultant electric field.

For a closed surface $\oint \vec{E} \cdot d\vec{s} = 0$,then:

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