Click to Chat
1800 2000 838
Complete Your Registration (Step 2 of 2 )
View complete IIT JEE section
View Complete Medical section
Medical Exam Calendar
View complete NRI section
View complete study material
Be the first one to answer
The answer is – mgd q (r 2 -d 2 ) 1/2
Click here to view the answer
sir not understand how three positive charges ??
Charge induced by +Q and -Q on the inner surface is -Q and +Q respectively. so, net charge induced is Zero. But the surface charge density is not zero inside. Region near to +Q is negatively charge...
When it is connected by wire, all charge from inside is conducted to the outer shell. And then potential inside the shell is a constant and equal to the potential at the surface i.e kQ/b kQ/a = V So,...
it will be but zero
all c/2 are connected in parallel , therefore answer is 2c
answer is a,c,d and final charge on plate 3 is Q/3
Force experience by the charge ‘Q’ in the electric field due to system of charges. As all the charges are positive hance there would be net zero field at centroid would be zero.
if this is a conductor then answer is sigma/(2*epsilon)
Charge is Charge density x dV and not dS
I am not going to give you a complete and lucarative solution, just go for the hints. 1. Increament in the force means, stifness produced in the ring due to radialy electrostatic force of repulsion....
F= 0.1 N = 9 * 10 9 * q 2 /r 2 q 2 = 0.1 * 0.01 2 /(9 * 10 9 ) q 2 = 10/9 * 10 -15 q= 3.34* 10 -8 no of electrons transferred = q/1.6 * 10 -19 = 3.34* 10 -8 /1.6 * 10 -19 =2 *10 11 (Approx.)