Question

A charge $Q$ is enclosed by a Gaussian spherical surface of radius $R$. If the radius is doubled, then the outward electric flux will

• A. increase four times
• B. be reduced to half
• C. remain the same
• D. be doubled

Answer 1

The Correct Option is C

To solve this question, we need to apply Gauss's law, which is a fundamental principle in electromagnetism. Gauss's law states that the total electric flux (\Phi) through a closed surface is equal to the charge enclosed (Q) divided by the permittivity of free space (\varepsilon_0):

\Phi = \frac{Q}{\varepsilon_0}

In this question, a charge Q is enclosed by a Gaussian spherical surface of radius R. The flux through this Gaussian surface is:

\Phi = \frac{Q}{\varepsilon_0}

Since the charge Q enclosed by the surface does not change when the radius is doubled, the total electric flux through the new surface remains the same. The electric flux is independent of the radius of the Gaussian surface as long as the enclosed charge remains constant.

Let's analyze the options:

1. Increase four times: Incorrect. The electric flux is not directly dependent on the radius of the Gaussian surface.
2. Be reduced to half: Incorrect. Reducing the radius does not affect the flux unless the enclosed charge changes.
3. Remain the same: Correct. The flux only depends on the enclosed charge Q, which remains unchanged.
4. Be doubled: Incorrect. Doubling the radius without changing the charge does not double the flux.

Therefore, the correct answer is that the outward electric flux will remain the same.