00:01
Hello students, here given that a is equal to inner radius 2 .4 mm and b is given as 3 .4 mm outer radius and the length is given as 2 mm and we know that epsilon 0 is equal to 8 .85 into 10 power minus 12 coulomb square newton inverse meter square.
00:27
Now to find the value of part a c by l the capacitance can be given for the radiuses having a and b which is given as c is equal to 2 pi epsilon 0 l divided by 2 .303 log b by a.
00:49
Therefore from this we can write it as c by l is equal to 2 pi epsilon 0 divided by 2 .303 log b by a.
01:08
Substituting the values we get that the value of c by l we get that here c by l is equal to 2 into 3 .14 into 8 .85 into 10 power minus 12 divided by 2 .303 log b is equal to 3 .4 a is equal to 2 .4 mm.
01:37
Therefore here it is 55 .578 into 10 power minus 12 divided by 2 .303 into here log 3 .4 by 2 .4 is equal to log 1 .416 .416.
01:56
Therefore by simplifying we get that c by l is equal to 55 .57 into 10 power minus 12 divided by 2 .303 into log 1 .416 is equals to 0 .3478.
02:17
Therefore c by l is equal to 0 .159 into 10 power minus 9 faraday or 0 .c by l is equal to 0 .159 picofaraday.
02:37
Now we will find the value for the b it is given that v is equal to 340 millivolts and we need to find the charge q1.
02:52
We know that q is equal to c into v from above expression c by l is given as the value of c by l is equal to 0 .159 into 10 power minus 9...