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Grade: 12
A point charge q is moving towards a grounded metallic sphere of radius R as shown in the figure with velocity V. The reading on the ammeter of the instant shown is
1) qRV / x^2
2) qV / R
3) qV / x
4) qR^2V / x^3
11 months ago

Answers : (2)

58 Points
in this question the sphere is metallic and is connected to the ground so potential of whole sphere = 0.
let charge accumelated on sphere be = q1 .
As potential at surface = 0. = k*q1/R + k*q/x. this gives q1 = -R*q/x so
to find current differentiate q1 with respec to x. here dx/dt = v
so you will finally get i = q*r*v/(x^2) (A)
2 months ago
Vikas TU
11498 Points
Dear student
Charge =q
distance =2R
radius =R
V= Potential at centre of the shell due to charge induced on shell.
As the electric field inside a conductor is zero. The electrostatic potential is constant it. The potential at centre will be the potential of entire conductor. The entire charge of conductor is on surface N.
Now, V= -1/4*pi*epselon *(q/R) 
The (−Ve) sign due to spherical shell of centre (inside).
2 months ago
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