In the diagram shown, electric flux over the surface of Gaussian surface (1) is 8×104 V−m and the flux over the Gaussian surface (2) is 12×104 V−m. Then the charge Q1 is equal to
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a
0.885 μC
b
1.475 μC
c
-0.56 μC
d
0.234 μC
answer is A.
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Detailed Solution
By Gauss's law, Q1−Q2=∈0 ϕ1 and Q1+Q2=∈0ϕ2∴ Q1=12∈0(ϕ1+ϕ2)=12×8.85×10−12×20×104C=0.885 μC
In the diagram shown, electric flux over the surface of Gaussian surface (1) is 8×104 V−m and the flux over the Gaussian surface (2) is 12×104 V−m. Then the charge Q1 is equal to