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

  • Thread starter mhrob24
  • Start date
30
2

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.
 

Answers and Replies

berkeman
Mentor
56,124
6,156
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? :smile:
 
30
2
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
Mentor
56,124
6,156
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). :smile:
 

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