Question:
If the radius of the spherical Gaussian surface is increased then the electric flux due to a point charge enclosed by the surface
If the radius of the spherical Gaussian surface is increased then the electric flux due to a point charge enclosed by the surface
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Electric flux through a closed surface depends only on enclosed charge, not on surface area or radius.
Updated On: Feb 18, 2026
- decreases
- remains unchanged
- increases
- is zero
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The Correct Option is B
Solution and Explanation
Step 1: Gauss’s law.
According to Gauss’s law, electric flux through a closed surface is \[ \Phi = \frac{q_{\text{enclosed}}}{\varepsilon_0}. \]
Step 2: Dependence on radius.
The flux depends only on the charge enclosed by the surface and not on the size or shape of the Gaussian surface.
Step 3: Conclusion.
Since the enclosed charge remains the same, the electric flux remains unchanged even if the radius increases.
According to Gauss’s law, electric flux through a closed surface is \[ \Phi = \frac{q_{\text{enclosed}}}{\varepsilon_0}. \]
Step 2: Dependence on radius.
The flux depends only on the charge enclosed by the surface and not on the size or shape of the Gaussian surface.
Step 3: Conclusion.
Since the enclosed charge remains the same, the electric flux remains unchanged even if the radius increases.
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