00:04
For problem 101, we are given two plates, a and b, with a is equal to negative 9 .5 micropoleum per square meter, and b is equal to 11 .6 micropoleum per square meter.
00:16
They are separated by 5cm distance and we are to treat them as infinite sheet.
00:22
So we are to get the net electric field for a, 4cm right of a.
00:29
So we have the point somewhere in the minimum.
00:32
Middle of them so this is a for b for cm to the left of plate a so this is b and for c for cm to the right of b so somewhere here so this is letter c so for this problem these distances we don't need them so our working equation would be e is equals to sigma divided by two epsilon not with e not is equal to 8 .85 times 10 raised to the negative 12.
01:11
So let's solve a.
01:13
So the effect of plate a on this point is an attraction and the plate b is a repulsion.
01:23
So this is ea eb.
01:26
So we have our net is equal to negative ea minus eb so we have negative sigma divided by 2 -epsilon minus sigma of a divided by 2 -epsilon so we have negative of 1 all over 2 times 8 .85 times 10 raised to the negative 12 multiplied by 9 .5 times 10 5 times 10 raised to 6 plus 11 .6 times 10 raised to 6 so our e is equal to 6.
02:07
So our e is equal to the negative 1 .19 times 10 raised to the 6 newton per column ihat...