00:01
Hi, here in this given problem radius of each metal sphere that is r is equal to 1 .1 centimeter charge over the first one that is 30 nanoculum and charge over the second one, q2, that is minus 20 nanoculum.
00:36
In the first part of the problem we have to find potential at the surface of both of them.
00:43
So for the first one v1 that will be given by k into q1 by r.
00:51
Means this is 9 into 10 to the power 9 multiplied by q1 which is 30 nanoculum or 30 multiplied by 10 to the par minus 9 column divided by radius 1 .1 centimeter or 1 .1 into 10 to the power minus 2 meter.
01:10
So this potential over the surface of the first metal sphere is calculated to be equal to 2 .45 into 10 to the power 4 volt.
01:23
Then over the surface of second sphere, k q2 by r means 9 into 10 to the power 9 into to minus 20 into 10th power minus 9, kulum divided by radius which is same as 1 .1 into 10 power minus 2.
01:43
So this v2 is calculated to be equal to minus 1 .64 into 10 raise to the power 4 volt.
01:56
Now in the second part of the problem, if we connect these two spheres.
02:04
With the help of a wire connecting, conducting wire on connecting them, the common potential that will be given by v is equal to total charge carried by them divided by common potential c1 plus c, sorry, common capacitance c1 plus c2, where c1 and c2 are the capacitances of individual capacitor, individual conductors, conducting spheres...