The flat base of a hemisphere of radius a wit | vedclass

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(B) According to Gauss's Law, for a closed surface with no charge inside, the net electric flux is zero.$\phi_{	ext{net}} = \phi_{	ext{curved}} + \p

Vedclass · Accepted Answer

$\frac{\pi a^2 E}{\sqrt{2}}$

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(B) According to Gauss's Law, for a closed surface with no charge inside, the net electric flux is zero.$\phi_{	ext{net}} = \phi_{	ext{curved}} + \phi_{	ext{base}} = 0$Therefore, $\phi_{	ext{curved}} = -\phi_{	ext{base}}$.The flux through the flat base is given by $\phi_{	ext{base}} = \vec{E} \cdot \vec{A}$, where $\vec{A}$ is the area vector of the base. The area vector $\vec{A}$ points vertically downwards (normal to the base), while the electric field $\vec{E}$ makes an angle of $45^{\circ}$ with the vertical.The angle between the electric field $\vec{E}$ and the area vector $\vec{A}$ is $180^{\circ} - 45^{\circ} = 135^{\circ}$.$\phi_{	ext{base}} = E A \cos(135^{\circ}) = E (\pi a^2) \left(-\frac{1}{\sqrt{2}}ight) = -\frac{\pi a^2 E}{\sqrt{2}}$.Since $\phi_{	ext{curved}} = -\phi_{	ext{base}}$, we get $\phi_{	ext{curved}} = \frac{\pi a^2 E}{\sqrt{2}}$.

The flat base of a hemisphere of radius $a$ with no charge inside it lies in a horizontal plane. $A$ uniform electric field $\vec{E}$ is applied at an angle $\frac{\pi}{4}$ with the vertical direction. The electric flux through the curved surface of the hemisphere is

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