00:01
Here, one conducting spherical cell is there.
00:07
Ok, so let's say this is the spherical cell.
00:13
Ok, this is the outer radius we can say and we can show here this is the inner radius.
00:22
Ok, so let's say this is r1 and this is r2.
00:30
Ok, so inner radius, inner radius r1 is given at how much? 7 centimeter.
00:44
That means 0 .07 meter.
00:47
Ok, outer radius, outer radius.
00:54
Ok, that is r2 is given at how much? 13 centimeter.
01:01
That means 0 .13 meter.
01:04
Ok, now q charged, q charged is there.
01:09
Ok, so q equal to it is given which is 35 nano coulomb.
01:17
That means we can write it is minus 35 into 10 to the power minus 9 coulomb, we can say.
01:25
Ok, now first is case one.
01:30
So i'm writing one.
01:32
So here, small r equal to 0 .5 r1.
01:38
That means 0 .5 into how much we can say it is 0 .07.
01:45
That is equal to 0 .035 meter.
01:50
Ok, now from from gauss law, what you can say? it is e dot ds, ok, integral equal to q enclosed by epsilon naught.
02:23
So we can write here e into, we can simply write e into 4 pi r square equal to q enclosed that is minus 35 into 10 to the power minus 9 by epsilon naught.
02:47
So electric field equal to what you can say, minus 35 into 10 to the power minus 9 by 4 pi epsilon naught into r square.
02:58
That is how much we can write 0 .035 whole square.
03:04
Ok, so we can write it is minus 9 into 10 to the power 9.
03:11
That is the value of 1 by 4 pi epsilon naught into 35 into 10 to the power minus 9 by 0 .035 whole square equal to, i'm getting the value as minus 2 .57 into 10 to the power 5 newton per coulomb.
03:33
So this is one answer.
03:37
Ok, now case two.
03:43
So here small r is 0 .5 times r1 plus r2...