# Find the charge on the plates of a parallel-plate capacitor w/ dielectric

## Homework Statement:

The area of each plate of a parallel plate capacitor is 0.024 m2. The plates are 2.75 mm apart with a dielectric material (κ = 3.0) between them. The maximum possible electric field between the plates is 3.25e5 V/m.

(a) What is the charge on the plates?

## Homework Equations:

C(dielectric) = Q/V(dielectric) , where V(dielectric) = 1/κ * V(original)

C(dielectric) = κ * C(original)

C = ε0 * (a/d) for parallel plate capacitor
This is an online HW question so maybe my digits are just off from rounding or something, but I don't know why I am not finding the correct answer. I got Q = 6.9e-8 as the magnitude of charge on each plate.

I basically just needed to calculate the original capacitance of this capacitor using c = ε0*(a/d) , then multiply this by the dielectric constant ( κ = 3.0) since the capacitance of a capacitor with a dielectric is just the original capacitance without the dielectric, multiplied by the dielectric constant (so capacitance can be increased by a factor of (κ) when a dielectric is inserted between its plates).

Area = .024 m^2
Distance between plates = 2.75mm

C(dielectric) = ε0 (.024/2.75) *3 = 2.31e-13

Now I need to find the voltage using V(dielectric) = (1/κ ) * V(original). We have a parallel plate capacitor which means our E field is uniform, therefore:

E = 3.25e5

V = E*D = 3.25e5 (2.75) = 893750

so, V(dielectric) = 893750(1/3) = 297916.6667

finally, we can use Q = C(dielectric) * V(dielectric) to get:

Q = 2.31e-13 (297916.6667) = 6.9e-8

I have no clue where I'm going wrong. I tried entering 7.0e-8 and 6.8e-8 incase it was a rounding issue and those were both incorrect as well. So either I'm just not working this problem correctly, or Webassign just sucks.

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berkeman
Mentor
Homework Statement: The area of each plate of a parallel plate capacitor is 0.024 m2. The plates are 2.75 mm apart with a dielectric material (κ = 3.0) between them. The maximum possible electric field between the plates is 3.25e5 V/m.

(a) What is the charge on the plates?
Homework Equations: C(dielectric) = Q/V(dielectric) , where V(dielectric) = 1/κ * V(original)

C(dielectric) = κ * C(original)

C = ε0 * (a/d) for parallel plate capacitor
It looks like you are double-correcting for the dielectric constant. Just include it in the Q=CV equation, and don't adjust your voltage for it.

Get the voltage from the maximum electric field and the plate separation, and the capacitance from the C = ε0 * εr (a/d) equation.

Can you try that and show that work? Yes, I think I see what you mean. I did get the correct answer (2.1e-7), but I want to make sure I know why I got the wrong answer at first

So excluding the fact that we have a dielectric :

C = ε0 * (a/d) = 7.72e-14

V = E*D = 893750 J/C

Now solve for Q, where Q = C*V, but enter the dielectric constant just once for either the voltage or the capacitance. I initially computed the voltage AND the capacitance with the dielectric constant which is wrong because I can also write C(dielectric) = (Q*κ)/V(original). That's how you derive the equation C(dielectric) = C(original) * κ. So I was factoring it in twice. So, :

Q = (C*κ) * V(original) = 2.1e-7

Thank you!

• berkeman
berkeman
Mentor
Q = (C*κ) * V(original) = 2.1e-7
I didn't follow what you did, but I got the same answer as you (just with a couple more sig figs). 