$(a)\;\large\frac{q(1-cos \theta)}{2 \in_{0}}\qquad(b)\;\large\frac{q sin \theta}{2\in_{0}} \qquad(c)\;\large\frac{q(1-cos \theta)}{ \in_{0}}\qquad(d)\;\large\frac{q sin \theta}{2 \in_{0}}$

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Answer : (a) $\;\large\frac{q(1-cos \theta)}{2 \in_{0}}$

Explanation :

Consider a Gaussian surface , a sphere with its centre at the top and radius the slant length of the cone . The flux through the whole sphere is $\;\large\frac{q}{\in_{0}}\;.$ Therefore the flux through the base of the cone is

$\phi_{e}=(\large\frac{S}{S_{0}})\;\large\frac{q}{\in_0{}}$

$S_{0}=\;$ area of whole sphere

S = area of sphere below the base of the cone .

Because all those field lines which pass through the base of the cone will pass through the cap of sphere

Let R= radius of Gaussion sphere

$S_{0}= \;$ area of whole sphere =$\; 4 \pi R^2$

S = area of sphere below the base of the cone which can be found by integration .

$dS = (2 \pi r )\;R\;d \alpha$

$dS=2 \pi R^2 sin \alpha \;d \alpha$

$S=\int_{0}^{\theta}\;dS =\int_{0}^{\theta}\;2 \pi R^2 sin \alpha \;d \alpha$

$=2 \pi R^2\;(1-cos \theta)$

The desired flux $\;\phi_{e} = \large\frac{S}{S_{0}}\;\large\frac{q}{\in_{0}}$

$=\large\frac{2 \pi R^2 (1-cos \theta ) }{4 \pi R^2}\;\large\frac{q}{\in_{0}}$

$\phi_{e} = \large\frac{q}{2 \in_{0}}\;(1-cos \theta)$

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