00:01
In the first part of this problem, we are going to calculate the electric field between the plates of the capacitor, that is e.
00:08
As the electric field between the plates of the capacitor is given by e equals to segment divided by epsilon 0, where this sigma is the choice density and it is written as a sigma equals to char it divided by area of the plates.
00:25
Now by inserting the value of this sigma into the equation, we can write it as e equals to charge, 2, q divided by a epsilon 0.
00:35
We call this equation as equation number 1.
00:38
Here this epsilon not is the permittivity of free space.
00:41
Now by inserting values into this equation, we can write it as e equals to.
00:46
We have the value for q as 1 .1, multiplied by eternalized power minus 6 coulamp divided by we have the value for area is 25 .25 .0 multiplied by 10x par minus 2 meter whole square into epsilon 0 is 8 .85 multiply by 10 10 x .m.
01:21
Square per newton meter square.
01:24
So this will give us the value for as e equals to 2 .0 multiplied by internal as power 6 volt per meter.
01:36
So this is the required answer for electric field.
01:45
Now in part b of this problem, we have to calculate the potential difference between the plates of the capacitor.
01:53
That is delta v...