Test Index

Electrostatics

Section: Physics

Question : 26 of 59

Marks: +1, -0

The electric flux passing through the curved surface of the hemisphere is $-\frac{Q}{2\varepsilon_{0}}\left(1-\frac{1}{\sqrt{2}}\right)$
The component of the electric field normal to the flat surface is constant over the surface
Total flux through the curved and the flat surfaces is $\frac{Q_{0}}{\varepsilon_{0}}$
The circumference of the flat surface is an equipotental

Solution:

Since the charge lies outside the sphere, net flux passing through the sphere is zero.

\phi_{\text{curved surface}}+\phi_{\text{disc}}=0

Option (C) is incorrect

\phi_{\text{curved surface}}=\phi_{\text{disc}}

\cos\theta=\frac{R}{\sqrt{R^{2}+r^{2}}}

E=\frac{1}{4\pi\varepsilon_{0}}\frac{Q}{R^{2}+r^{2}}

\phi_{\text{disc}}=\overline{\int E}\cdot\overline{dA}

=\int\left(\frac{1}{4\pi\varepsilon_{0}}\frac{Q}{R^{2}+r^{2}}\times2\pi r\,dr\times\cos\theta\right)

=\frac{Q\cdot2\pi}{4\pi\varepsilon_{0}}\int\frac{r\,dr}{R^{2}+r^{2}}\times\frac{R}{(R^{2}+r^{2})^{1/2}}

\frac{QR}{2\varepsilon_{0}}\int\limits_{0}^{R}\frac{r\,dr}{(R^{2}+r^{2})^{3/2}}

=\frac{QR}{2\varepsilon_{0}}\left[\frac{1}{2}\frac{(R^{2}+r^{2})^{-1/2}}{-1/2}\right]_{0}^{R}

=\frac{QR}{2\varepsilon_{0}}\left[\frac{1}{R}-\frac{1}{\sqrt{R^{2}+R^{2}}}\right]

=\frac{Q}{2\varepsilon_{0}}\left[1-\frac{1}{\sqrt{2}}\right]

\Rightarrow\phi_{\text{curved surface}}=-\frac{Q}{2\varepsilon_{0}}

Option (A) is correct Potential at any point on the circumference of the flat surface is

\frac{1}{4\pi\varepsilon_{0}}\frac{Q}{\sqrt{R^{2}+R^{2}}}

=\frac{Q}{4\pi\varepsilon_{0}(\sqrt{2}R)}

Hence it is equipotential

Option (D) is correct

E=\frac{1}{4\pi\varepsilon_{0}}\frac{Q}{\left(\frac{R}{\cos\theta}\right)^{2}}=\frac{\theta}{4\pi\varepsilon_{0}R^{2}}\cos^{2}\theta

E_{\text{normal}}=E\cos\theta=\frac{Q}{4\pi\varepsilon_{0}R^{2}}\cos^{3}\theta

Which is not constant

Option (B) is in correct

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