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Refer to the arrangement of charges in F...

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

A

total flux through the surface of the sphere is `(-Q)/(epsi_(0))`

B

field on the surface of the sphere is `(-Q)/(4piepsi_(0)R^(2))`

C

flux through the surface of sphere due to 5Q is zero

D

field on the surface of sphere due to -2Q is same everywhere

Text Solution

Verified by Experts

The correct Answer is:
A, C
Gauss' law states that total flux of an enclosed surface is given by `(q)/(epsi_(0))` where q is the charge enclosed by the surface. Thus, from figures,
Total charge inside the surface is `=Q-2Q=-Q`
`therefore` Total flux through the surface of the sphere`=(-Q)/(epsi_(0))`
Now, considering charge 5Q. Charge 5Q lies outside the surcae, thus it makes no contribution to electric flux through the given surface.
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