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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By Gauss law the flux depends only on enclosed charge.
Updated On: Oct 1, 2026
- increases
- remains unchanged
- is zero
- decreases
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The Correct Option is B
Solution and Explanation
Step 1: Understanding the Concept:
Gauss's law states that the net electric flux through a closed surface is \(\Phi=\dfrac{q_{enclosed}}{\varepsilon_0}\).
Step 2: Apply:
The point charge stays inside when the sphere is enlarged, so \(q_{enclosed}\) does not change. The flux remains \(\dfrac{q}{\varepsilon_0}\).
Step 3: Why the other options are wrong.
The field falls as \(1/r^2\), but the area grows as \(r^2\), so their product is constant. Flux does not rise, fall, or become zero.
Final Answer:
The flux remains unchanged. \[ \boxed{\text{(B) }\text{remains unchanged}} \]
Gauss's law states that the net electric flux through a closed surface is \(\Phi=\dfrac{q_{enclosed}}{\varepsilon_0}\).
Step 2: Apply:
The point charge stays inside when the sphere is enlarged, so \(q_{enclosed}\) does not change. The flux remains \(\dfrac{q}{\varepsilon_0}\).
Step 3: Why the other options are wrong.
The field falls as \(1/r^2\), but the area grows as \(r^2\), so their product is constant. Flux does not rise, fall, or become zero.
Final Answer:
The flux remains unchanged. \[ \boxed{\text{(B) }\text{remains unchanged}} \]
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