Question

The electric field inside a uniformly charged spherical shell of radius R is ________.

• A. directly proportional to the charge within the shell
• B. inversely proportional to $R^2$
• C. same as that outside the shell
• D. zero
• E. maximum at the centre

Hint:

Electrostatic shielding: Inside a conductor or a shell, the electric field is always zero.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept
This is a classic problem in electrostatics that is solved using Gauss's Law. A spherical shell has all of its charge residing on its surface. We want to find the electric field at any point \textit{inside} the cavity of the shell.
Step 2: Key Formula or Approach
Gauss's Law states that the net electric flux (\(\Phi_E\)) through any closed surface (called a Gaussian surface) is equal to the net charge enclosed (\(q_{enc}\)) by the surface, divided by the permittivity of free space (\(\epsilon_0\)).
\[ \oint \vec{E} \cdot d\vec{A} = \frac{q_{enc}}{\epsilon_0} \] To find the field inside the shell, we construct a spherical Gaussian surface with a radius \(r<R\) concentric with the shell.
Step 3: Detailed Explanation
1. Construct a Gaussian Surface.
Consider a point P inside the uniformly charged spherical shell, at a distance \(r\) from the center, where \(r<R\). We draw a spherical Gaussian surface of radius \(r\) passing through P.
2. Apply Gauss's Law.
- We need to find the total charge enclosed by this Gaussian surface, \(q_{enc}\).
- The problem states it is a "spherical shell", which means all the charge resides on the surface of the sphere of radius R.
- Since our Gaussian surface has a radius \(r<R\), it is entirely inside the shell and does not enclose any of the charge.
- Therefore, the enclosed charge \(q_{enc} = 0\).
3. Calculate the Electric Field.
Substituting \(q_{enc} = 0\) into Gauss's Law:
\[ \oint \vec{E} \cdot d\vec{A} = \frac{0}{\epsilon_0} = 0 \] This means the net electric flux through our Gaussian surface is zero.
Due to the spherical symmetry of the problem, if the electric field \(\vec{E}\) were non-zero, it would have to be constant in magnitude and directed radially at every point on the Gaussian surface. In that case, the flux integral would be \(E \cdot (4\pi r^2)\).
So, \(E \cdot (4\pi r^2) = 0\).
Since the area \(4\pi r^2\) is not zero, the electric field magnitude \(E\) must be zero.
\[ E = 0 \] This holds true for any point inside the shell (\(0 \le r<R\)).
Step 4: Final Answer
The electric field inside a uniformly charged spherical shell is zero everywhere.