Calculating the capacitance problem

[ctpengage]

Homework Statement

If an uncharged parallel-plate capacitor (capacitance C) is connected to a battery, one plate becomes negatively charged as electrons move to the plate face (area A). The depth d from which the electrons come in the plate in a particular capacitor is plotted against a range of values for the potential difference V of the battery. When d is one picometre, the potential difference is 20V. The gradient is equal to 5x10^-14. What is the ratio C/A.

Homework Equations

q=ε₀EA
q=CV

The Attempt at a Solution

I combined the above two equations to get the expressions C/A=ε₀/D
However that doesn't provide me with a numerical solution for the ratio C/A.

Can anyone please help?

 

[Redbelly98]

I don't understand what is meant by "the depth ... from which the electrons come in the plate".

Also, what is this "gradient equal to 5x10^-14" mean? And what are the units?