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?

 

[baba89]

The capacitance of a parallel-plate capacitor is given by the equation C = ε₀A/d, where ε₀ is the permittivity of free space, A is the area of the plates, and d is the distance between the plates. In this problem, we are given the potential difference V and the distance d, and we need to find the ratio C/A.

First, we can rearrange the equation for capacitance to solve for A: A = Cd/ε₀. Then, we can substitute this expression for A into the ratio C/A to get:

C/A = C / (Cd/ε₀) = ε₀ / d

Next, we can use the given information to solve for the permittivity of free space, ε₀. We know that when d = 1 picometre, V = 20V, and the gradient is equal to 5x10^-14. Using the equation for potential difference (V = Ed), we can solve for the electric field E:

E = V/d = 20V / (1x10^-12 m) = 2x10^13 V/m

Then, using the equation for electric field (E = σ/ε₀), we can solve for the surface charge density σ:

σ = ε₀E = (5x10^-14) x (2x10^13) = 1x10^-13 C/m^2

Finally, we can use the equation for surface charge density (σ = q/A) to solve for the charge q on one of the plates:

q = σA = (1x10^-13 C/m^2) x A

We can then substitute this expression for q into the equation for capacitance (C = q/V) to get:

C = (1x10^-13 C/m^2) x A / 20V = (1/20) x (1x10^-13 C/m^2) x A

Now, we can substitute this expression for C into the ratio C/A to get:

C/A = [(1/20) x (1x10^-13 C/m^2) x A] / A = (1/20) x (1x10^-13 C/m^2) = 5x10^-15 F/m^2

Therefore, the ratio C/A is equal to 5x10^-15 F/m^2.