Gauss's Law
A spherical Gaussian surface of radius 15 cm is drawn around a point charge. How does the total electric flux through this surface change if the radius is increased to 30 cm?
Select the correct option:
Solution
It remains unchanged
Gauss's Law states that the total electric flux through any closed surface is equal to the net charge enclosed divided by the permittivity of free space: (\Phi_E = \frac{Q_{enc}}{\varepsilon_0}). This is one of the most fundamental results in electrostatics and holds regardless of the size or shape of the Gaussian surface. The flux depends only on the net charge enclosed, not on the shape, size, or position of the surface. Since both surfaces (radius 15 cm and radius 30 cm) enclose the same point charge (Q), the total flux through each is (\Phi = Q/\varepsilon_0). Option 'It doubles' is incorrect because increasing radius does not change the enclosed charge. Option 'It halves' is incorrect because the flux is independent of surface size for the same enclosed charge. Option 'It increases four times' is incorrect because this would be the change in surface area, but flux is not proportional to area alone. This is a direct JEE Main application of Gauss's Law and is commonly tested as a conceptual question. Plausibility check: all field lines emanating from the charge must pass through any surface surrounding it, confirming constant flux.
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About This Question
- Subject
- physics
- Chapter
- electrostatics
- Topic
- gauss's law
- Difficulty
- Easy
- Year
- 2025
Solution
Correct Answer:
It remains unchanged
Gauss's Law states that the total electric flux through any closed surface is equal to the net charge enclosed divided by the permittivity of free space: (\Phi_E = \frac{Q_{enc}}{\varepsilon_0}). This is one of the most fundamental results in electrostatics and holds regardless of the size or shape of the Gaussian surface. The flux depends only on the net charge enclosed, not on the shape, size, or position of the surface. Since both surfaces (radius 15 cm and radius 30 cm) enclose the same point charge (Q), the total flux through each is (\Phi = Q/\varepsilon_0). Option 'It doubles' is incorrect because increasing radius does not change the enclosed charge. Option 'It halves' is incorrect because the flux is independent of surface size for the same enclosed charge. Option 'It increases four times' is incorrect because this would be the change in surface area, but flux is not proportional to area alone. This is a direct JEE Main application of Gauss's Law and is commonly tested as a conceptual question. Plausibility check: all field lines emanating from the charge must pass through any surface surrounding it, confirming constant flux.
This easy difficulty physics question is from the chapter electrostatics, covering the topic of gauss's law. It appeared in the 2025 exam.
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