Question

A charge \(Q\) is enclosed by a spherical Gaussian surface of radius \(R\). If the radius is doubled, then the total electric flux through the surface:

• A. becomes four times
• B. remains the same
• C. becomes half
• D. becomes twice

Hint:

By Gauss law the flux equals the enclosed charge over \(\varepsilon_0\). The enclosed charge is the same, so the flux is the same.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Gauss law says the total electric flux through any closed surface equals the charge enclosed divided by \(\varepsilon_0\).

Step 2: Key Formula or Approach:
\[ \Phi = \frac{Q_{enclosed}}{\varepsilon_0} \]
The flux depends only on the enclosed charge, not on the size or shape of the surface.

Step 3: Detailed Explanation:
For the radius \(R\), the enclosed charge is \(Q\), so \(\Phi_1 = Q/\varepsilon_0\).
When the radius becomes \(2R\), the same charge \(Q\) is still the only charge inside. So \(\Phi_2 = Q/\varepsilon_0\).
The ratio is \(\Phi_2 / \Phi_1 = 1\).

Step 4: Why the other options are wrong:
Option (A), four times: this uses the area \(4\pi r^2\) alone and forgets that the field becomes \(1/4\) at the larger radius.
Option (C), half, and option (D), twice: no step of Gauss law gives such a change, because the enclosed charge has not changed.

Final Answer:
The total flux stays the same. This is option (B). \[ \boxed{\text{(B) remains the same}} \]