00:01
Hi students, here in this question the potential is given as v is equal to 2xy -3xz plus 5y square.
00:11
Now we have to find out the potential at 3 ,0 ,1.
00:18
V at 3 ,0 ,1 that means x equal to 3, y is equal to 0, z is equal to 1.
00:27
X equal to 3, y is equal to 0, z is equal to 1.
00:31
So what will be v at 3 ,0 ,1 just substituting the values for x and y and z.
00:38
Okay, so 2 into 3 into 2 into 3 into 0 minus 3 into 1 into 3 plus 5 into 0 square.
00:56
So that's equal to minus 9 volt.
01:00
So this is v at minus 3 ,0 ,1.
01:07
Therefore potential difference at the point 3 ,0 ,1 is v is equal to minus 9 volt.
01:26
Now we have to find out the components of the electric field, ex, ey and ez.
01:33
So we know that e, e is nothing but minus del v.
01:42
So what is del v? so minus of dou v by dou x into i cap plus dou v by dou y into j cap plus dou v by dou z into k cap.
02:00
So that's equal to, here we can substitute the values.
02:05
Okay, so what is v? v is 2xy minus 3zx plus 5y square.
02:11
So we have to partially differentiate with respect to x, y and z.
02:16
Okay, so here it will be dou v by dou x.
02:22
So dou v by dou x 2y into 1 minus 3z into 1.
02:30
So it is 2y minus 3z into i cap.
02:36
Okay, minus of minus, minus of plus.
02:41
So it will be minus...