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Question
Refer to the arrangement of charges in figure and a Gaussian surface of radius R with Q at the centre. Then
- total flux through the surface of the sphere is `(-Q)/ε_0`.
- field on the surface of the sphere is `(-Q)/(4 piε_0 R^2)`.
- flux through the surface of sphere due to 5Q is zero.
- field on the surface of sphere due to –2Q is same everywhere.
Options
a and d
a and c
b and d
c and d
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Solution
a and c
Explanation:
From Gauss' law, we know `oint_s vecE * dvecS = q_(enclosed)/ε_0`.
Thus, from figure, Total charge inside the Gaussian surface `q_(enclosed)` = Q – 2Q = – Q
The charge 5Q lies outside the surface, thus it makes no contribution to electric flux through the given surface.
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