Rankers Physics

Electric Flux: Practice Problem & Solution

The electric flux Φ through a hemisphere surface of radius R, placed in a uniform electric field of intensity E parallel to the axis of its circular plane is:
\[2\pi RE\]
\[2\pi R^{2}E\]
\[\pi R^{2}E\]
\[\left( 4/3 \right)\pi R^{3}E\]

Solution Explained:

To solve this problem, we apply the core principles of Electric Flux. Understanding the underlying formula is key to arriving at the correct answer below:

The electric flux through the hemisphere is due to the uniform electric field passing perpendicularly through the circular plane of radius \( R \).

Flux through the circular plane:
\[
\Phi = E \cdot \text{Area of circular plane} = E \cdot \pi R^2
\]

Since no flux passes through the curved surface (field is parallel to it), the total flux is:
\[
\Phi = \pi R^2 E
\]

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Similar Practice Questions

Explore conceptually related problems to test your knowledge.

1. The electric field in a region is directed outward and is proportional to the distance \(r\) from the origin. If we take a spherical volume of radius \(r\) taking centre at the origin, then the charge contained inside the volume is proportional to
2. It is required to hold equal charges q in equilibrium at the corners of a square. What charge when placed at the centre of the square will do this ?
3. Two surfaces S1 & S2 are shown in figure. when a pt charge is placed at pt P Flux associated with S1 is Φ1 & S2 is Φ2. Which is correct.
4. A cylinder of radius R and length L is placed in a uniform electric field E parallel to the cylinder axis. The total flux from the surface of the cylinder is:
5. The length of each side of a cubical closed surface is l. If charge q is situated on one of the vertices of the cube, then find the flux passing through shaded face of the cube.