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  • Refer to the arrangement of charges in Fig. 1.6 and a Gaussian surface of radius R with Q at the centre. Then

Refer to the arrangement of charges in Fig. 1.6 and a Gaussian surface of radius R with Q at the centre. Then

A. total flux through the surface of the sphere is \frac{-Q}{\varepsilon _0}

B. field on the surface of the sphere is \frac{-Q}{4\pi \varepsilon _0R^{2}}.
C. flux through the surface of the sphere due to 5Q is zero.
D. field on the surface of the sphere due to –2Q is the same everywhere.

Answers (1)

The correct answers are the options a,c

Explanation:-

Since Q is at the center of the Gaussian surface (a sphere with radius R), the total charge enclosed by the surface is Q. If 5Q is outside the Gaussian surface, it does not contribute to the flux through the surface. The flux only depends on the charge enclosed by the Gaussian surface. Hence making no contribution to the electric flux. So, a and c are correct options.

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infoexpert21

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