The flat base of a hemisphere of radius a with no charge inside it lies in a horizontal plane. A uniform…

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
  1. $\pi a^2 E$
  2. $\frac{\pi a^2 E}{\sqrt{2}}$
  3. $\frac{\pi a^2 E}{2 \sqrt{2}}$
  4. $\frac{\left(\pi+2 \pi \phi^2 E\right.}{\left(2 \sqrt{2}{ }^2\right)}$

Solution

We know that, $ \phi=\oint E \cdot d S=E \oint d S \cos 45^{\circ} $ In case of hemisphere $ \phi_{\text {curved }}=\phi_{\text {circular }} $ Therefore, $\phi_{\text {curved }}=E \pi a^2 \cdot \frac{1}{\sqrt{2}}=\frac{E \pi a^2}{\sqrt{2}}$

Asked in: JEE Main 2012 (19 May Online)

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