Join now for JEE/NEET and also prepare for Boards Join now for JEE/NEET and also prepare for Boards. Register Now
Hey there! We receieved your request
Stay Tuned as we are going to contact you within 1 Hour
One of our academic counsellors will contact you within 1 working day.
Click to Chat
1800-1023-196
+91-120-4616500
CART 0
Use Coupon: CART20 and get 20% off on all online Study Material
Welcome User
OR
LOGIN
Complete Your Registration (Step 2 of 2 )
Sit and relax as our customer representative will contact you within 1 business day
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 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).
Dear , Preparing for entrance exams? Register yourself for the free demo class from askiitians.
points won -