Question

Calculate the electric field at a point due to a uniformly charged spherical shell.

• A. \( \frac{Q}{4 \pi \epsilon_0 r^2} \)
• B. \( 0 \)
• C. \( \frac{Q}{4 \pi \epsilon_0 r} \)
• D. \( \frac{Q}{8 \pi \epsilon_0 r^2} \)

Hint:

For spherical symmetry, remember that the electric field inside a uniformly charged spherical shell is always zero, regardless of the shell’s charge or radius.

Answer 1

The Correct Option is B

We are asked to find the electric field at a point due to a uniformly charged spherical shell. The solution involves understanding the application of Gauss's Law, which is a fundamental concept in electrostatics.
1. Gauss's Law states that the electric flux through a closed surface is proportional to the charge enclosed by that surface: \[ \oint \vec{E} \cdot d\vec{A} = \frac{Q_{\text{enc}}}{\epsilon_0} \] Where: - \( \vec{E} \) is the electric field, - \( d\vec{A} \) is an infinitesimal area element, - \( Q_{\text{enc}} \) is the enclosed charge, - \( \epsilon_0 \) is the permittivity of free space.
2. For a spherical shell, the symmetry of the charge distribution means that the electric field inside the shell must be uniform and radially symmetric. - At any point inside the spherical shell, the contribution to the electric field from all parts of the shell cancels out due to symmetry. - This is because the field due to a charge element at one part of the shell is canceled by the field due to a charge element at the opposite side of the shell. Thus, the electric field inside the shell is zero.
3. Therefore, the electric field at a point inside the shell is zero. This result is independent of the radius of the shell or the charge enclosed by the shell.
4. Outside the shell (for \( r < R \)), the electric field behaves as though the entire charge is concentrated at the center, but the question asks for the field at a point inside the shell. Thus, option (1), (3), and (4) are incorrect.
Option (2) is correct because the electric field inside the shell is zero.