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

• mhrob24
In summary, this HW question asks for the magnitude of charge on each plate of a parallel plate capacitor. Using the equation C(dielectric) = Q/V(dielectric), the charge on the plates is 6.9e-8.
mhrob24
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?
Relevant 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.

mhrob24 said:
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
mhrob24 said:
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).

1. What is a parallel-plate capacitor with dielectric?

A parallel-plate capacitor with dielectric is an electrical component that consists of two parallel plates separated by a dielectric material, such as air or plastic. It is used to store electrical energy and is commonly found in electronic devices.

2. How do you find the charge on the plates of a parallel-plate capacitor with dielectric?

The charge on the plates of a parallel-plate capacitor with dielectric can be found by using the formula Q = CV, where Q is the charge, C is the capacitance, and V is the voltage. The capacitance can be calculated by using the formula C = εA/d, where ε is the permittivity of the dielectric, A is the area of the plates, and d is the distance between the plates.

3. What is the role of the dielectric in a parallel-plate capacitor?

The dielectric in a parallel-plate capacitor serves to increase the capacitance by reducing the electric field between the plates. This allows for a larger charge to be stored on the plates for a given voltage. It also serves to prevent the plates from coming into contact with each other, which could cause a short circuit.

4. How does the dielectric constant affect the charge on the plates?

The dielectric constant, also known as the relative permittivity, is a measure of how well a material can store electrical energy. The higher the dielectric constant, the greater the charge that can be stored on the plates. This is because a higher dielectric constant means a higher capacitance, which results in a higher charge for a given voltage.

5. How does the distance between the plates affect the charge on a parallel-plate capacitor with dielectric?

The distance between the plates, also known as the plate separation, plays a significant role in determining the charge on a parallel-plate capacitor with dielectric. As the distance between the plates decreases, the capacitance increases, resulting in a higher charge for a given voltage. This is due to the fact that the electric field between the plates is stronger when the plates are closer together, allowing for more charge to be stored.

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