Question

A hemisphere of radius $R$ is placed in a uniform electric field $E$ so that its axis is parallel to the field. Which of the following statement(s) is/are true?

• A. Flux through the curved surface of hemisphere is πR2E.
• B. Flux through the circular surface of hemisphere is πR2E.
• C. Total flux enclosed is zero.
• D. Work done in moving a point charge q from A to B via the path ACB depends upon R.

Answer 1

The Correct Option is A, C

Step 1: Understand Electric Flux

Electric flux, \(\Phi\), through a surface is defined as: \[\Phi = \int \vec{E} \cdot d\vec{A}\] For a uniform electric field and a flat surface, this simplifies to: \[\Phi = E A \cos{\theta}\] where \(A\) is the area of the surface and \(\theta\) is the angle between the electric field and the normal vector to the surface.

Step 2: Flux Through the Circular Surface

The circular surface is perpendicular to the electric field. Therefore, the angle \(\theta\) between the electric field and the normal to the surface is 180 degrees (or 0 degrees, depending on which direction you define the normal to point). The area of the circular surface is \(A = \pi R^2\). The flux through the circular surface is: \[\Phi_{\text{circular}} = E (\pi R^2) \cos{180^\circ} = -E \pi R^2\] The negative sign indicates that the flux is entering the hemisphere.

Step 3: Flux Through the Curved Surface

Since there is no charge enclosed within the hemisphere, according to Gauss's Law, the total flux through the closed surface (hemisphere) must be zero. \[\Phi_{\text{total}} = \Phi_{\text{circular}} + \Phi_{\text{curved}} = 0\] Therefore, the flux through the curved surface must be equal in magnitude but opposite in sign to the flux through the circular surface: \[\Phi_{\text{curved}} = - \Phi_{\text{circular}} = E \pi R^2\] This is the outgoing flux.

Step 4: Total Flux Enclosed

As mentioned in Step 3, Gauss's Law states that the total electric flux through a closed surface is proportional to the enclosed charge. Since there is no charge enclosed within the hemisphere, the total flux enclosed is zero: \[\Phi_{\text{total}} = 0\]

Step 5: Work Done in Moving a Charge

The work done in moving a point charge \(q\) from point \(A\) to point \(B\) in an electric field is given by: \[W = -q \Delta V = -q (V_B - V_A)\] where \(\Delta V\) is the potential difference between points \(A\) and \(B\). In a uniform electric field, the potential difference only depends on the displacement along the direction of the electric field. Since points A and B are at the same height, the electric potential difference is independent of R. Therefore, the work done depends on the potential difference between points \(A\) and \(B\), and it is independent of the path taken (since the electric field is conservative) and also independent of \(R\).

Step 6: Determine the Correct Statements

1. Flux through the curved surface of hemisphere is πR²E: This statement is TRUE.
2. Flux through the circular surface of hemisphere is πR²E: This statement is FALSE; the flux is -πR²E.
3. Total flux enclosed is zero: This statement is TRUE.
4. Work done in moving a point charge q from A to B via the path ACB depends upon R: This statement is FALSE.

Conclusion

The correct statements are:

• Flux through the curved surface of the hemisphere is πR²E.
• Total flux enclosed is zero.