A point charge is placed just outside an imaginary hemispherical surface of radius as shown in the figure. Which of the following statements is/are correct? [2017]

Question

A point charge $+ Q$ is placed just outside an imaginary hemispherical surface of radius $R$ as shown in the figure. Which of the following statements is/are correct? [2017]

Smart Achievers · Accepted Answer

(1, 4)

The circumference of the flat surface is an equipotential $V = \frac{K Q}{\sqrt{2} R}$

because the circumference is equidistant from $+ Q$ . The component of electric field perpendicular to the flat surface is $E \cos θ .$

Here E as well as $θ$ changes for different points on the flat surface. The total flux through the curved and flat surfaces should be less than $\frac{Q}{ε_{0}} .$

The solid angle subtended by the flat surface at $P$

$= 2 π (1 - \cos θ) = 2 π (1 - \cos 45 °)$ $= 2 π (1 - \frac{1}{\sqrt{2}})$

$∴$ Flux passing through the curved surface

$= - \frac{Q}{ε_{0}} \cdot \frac{2 π (1 - \frac{1}{\sqrt{2}})}{4 π}$ $= - \frac{Q}{2 ε_{0}} (1 - \frac{1}{\sqrt{2}})$

Solution

Q.
A point charge $+ Q$ is placed just outside an imaginary hemispherical surface of radius $R$ as shown in the figure. Which of the following statements is/are correct [2017]

1 The electric flux passing through the curved surface of the hemisphere is $- \frac{Q}{2 ε_{0}} (1 - \frac{1}{\sqrt{2}})$

2 Total flux through the curved and the flat surfaces is $\frac{Q}{ε_{0}}$

3 The component of the electric field normal to the flat surface is constant over the surface.

4 The circumference of the flat surface is an equipotential.

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