A battery V=20Vis connected across a parallel plate capacitor, each plate with area 2.5x10^-2 m^2, separation distance 6x10^-14. (You can assume it's approximately infinite plates.) Position b is located 3x10^-4 m above position a, and position c lies 6x10^-4 m to the right of b. All 3 of the positions are between the plates and position a is touching one of the plates.
a) Obtain the voltage difference between points a and b, a and c, and between b and c. b)Obtain the charge on one of the plates.

a)

Since a and b are on the same plate (a simple sketch will confirm this c is 6x10^-14 from b, which is also the distance of separation of the plates, meaning that they b touches the left plate and c touches the right plate), they are on the same potential.

Hence, the potential difference between points a and b is zero. (the difference in their potential is zero since they have the same potential!)

a and c are on different plates. The plates themselves have a potential difference of 20 V across them, so, this is the potential difference between those plates. Therefore, the potential difference between the points a and b is 20 V.

Now, I think you can do the third part, for the potential difference between b and c. :)

b)

The charge is given by:

Q = CV

Now, C = \frac{\epsilon A}{d}

Where epsilon is the permittivity of the insulator,
A is the area of the plate,
d is the separation between the two plates.

You need to have the permittivity constant for the insulator between the two capacitor plates. If you are told to use the permittivity of free space, then the value is 8.85\times 10^{-12}\ Fm^{-1}.

I hope it helped! Post your answer! :)