When a potential difference of 180 V is applied to the plates of a parallel-plate capacitor, the plates carry a surface charge density of 29.0 nC/cm^{2}. What is the spacing between the plates?

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+1 vote

Best answer

First establish that the formula for the capacitor of a parallel plate is:

C = E * (A/d)

where E = permittivity of free space = 8.854*10^-12 C^2/N*m^2

A = area, and d = distance between the plates.

Now also use the formula for capacitance, which is

C = Q/V

Q = charge

V = Voltage or potential difference = 180 V

Rearrange the equation for capacitance so that:

Q = C*V

now plug back in the formulat for parallel plate Capacitor for C, resulting in

Q = E * (A/d) * V

In the question you are given charge density of 29 nC/cm^2, first convert this to SI units.

so = 29 nC/cm^2 = 29*10^-6 C/ (1*10^-2 m)^2

note that charge density is equal to Q/A

so Q/A = 29*10^-6 C/ (1*10^-2 m)^2

So rearrange this equation: Q = E * (A/d) * V, so that Q/A are on the left:

Q/A = (E/d) * V

and so, charge density = (E/d) * V

Now rearrange equation and solve for d.

d = E*V / charge density

Substitute in values and solve for d:

d = (8.854*10^-12 C^2/N*m^2) * (180V) / 29*10^-6 C/ (1*10^-2 m)^2

d = 5.49 * 10 ^-9 m = 5.49 µm

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