A hemispherical surface of radius a is placed | vedclass

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Question

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

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) According to Gauss's Law,for a closed surface with no net charge enclosed,the total electric flux is zero.$\phi_{	ext{total}} = \phi_{	ext{curved}} + \phi_{	ext{flat}} = 0$Therefore,$\phi_{	ext{curved}} = -\phi_{	ext{flat}}$.The flux through the flat circular surface is given by $\phi_{	ext{flat}} = \vec{E} \cdot \vec{A} = EA \cos(	heta)$,where $	heta$ is the angle between the electric field and the area vector.The area vector of the flat surface is vertical. The electric field makes an angle of $45^{\circ}$ with the vertical. Thus,the angle between the electric field and the area vector is $45^{\circ}$.$\phi_{	ext{flat}} = E(\pi a^2) \cos(45^{\circ}) = E(\pi a^2) \frac{1}{\sqrt{2}}$.Since the field lines enter through the flat surface and exit through the curved surface,the magnitude of the flux through the curved surface is equal to the magnitude of the flux through the flat surface.$\phi_{	ext{curved}} = \frac{\pi a^2 E}{\sqrt{2}}$.

$A$ hemispherical surface of radius $a$ is placed in a uniform electric field $E$ such that the flat surface makes an angle of $\pi /4$ with the vertical. There is no charge inside the hemisphere. The electric flux passing through the curved surface of the hemisphere is:

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