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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 is1) qRV / x^22) qV / R3) qV / x4) qR^2V / x^3
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 soto find current differentiate q1 with respec to x. here dx/dt = vso you will finally get i = q*r*v/(x^2) (A)
Dear studentCharge =qdistance =2Rradius =RV= 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).
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