Question:
Four closed surfaces and corresponding charge distributions are shown below.
Let the respective electric fluxes through the surfaces be \(\Phi_1\), \(\Phi_2\), \(\Phi_3\) and \(\Phi_4\). Then :
Four closed surfaces and corresponding charge distributions are shown below.
Let the respective electric fluxes through the surfaces be \(\Phi_1\), \(\Phi_2\), \(\Phi_3\) and \(\Phi_4\). Then :
Let the respective electric fluxes through the surfaces be \(\Phi_1\), \(\Phi_2\), \(\Phi_3\) and \(\Phi_4\). Then :
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By Gauss's law the flux depends only on the net charge enclosed; charges outside a surface do not contribute.
Updated On: Nov 11, 2026
- \(\Phi_1 \lt \Phi_2 = \Phi_3 \gt \Phi_4\)
- \(\Phi_1 \gt \Phi_2 \gt \Phi_3 \gt \Phi_4\)
- \(\Phi_1 = \Phi_2 = \Phi_3 = \Phi_4\)
- \(\Phi_1 \gt \Phi_3\,;\ \Phi_2 \lt \Phi_4\)
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The Correct Option is C
Solution and Explanation
Gauss's law: \(\Phi = \frac{q_{enc}}{\varepsilon_0}\), charges outside contribute zero net flux.
S1: encloses \(2q\) so \(q_{enc} = 2q\).
S2: encloses \(q, -q, q, q\) so \(q_{enc} = 2q\).
S3: encloses \(q, q\) (the \(5q\) is outside) so \(q_{enc} = 2q\).
S4: encloses \(8q, -2q, -4q\) (the \(3q\) is outside) so \(q_{enc} = 8q - 2q - 4q = 2q\).
All four fluxes equal \(\frac{2q}{\varepsilon_0}\): \(\Phi_1 = \Phi_2 = \Phi_3 = \Phi_4\).
Answer: (3) \(\Phi_1 = \Phi_2 = \Phi_3 = \Phi_4\)
S1: encloses \(2q\) so \(q_{enc} = 2q\).
S2: encloses \(q, -q, q, q\) so \(q_{enc} = 2q\).
S3: encloses \(q, q\) (the \(5q\) is outside) so \(q_{enc} = 2q\).
S4: encloses \(8q, -2q, -4q\) (the \(3q\) is outside) so \(q_{enc} = 8q - 2q - 4q = 2q\).
All four fluxes equal \(\frac{2q}{\varepsilon_0}\): \(\Phi_1 = \Phi_2 = \Phi_3 = \Phi_4\).
Answer: (3) \(\Phi_1 = \Phi_2 = \Phi_3 = \Phi_4\)
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