Dielectric Problem: Q&A to Calculate Electric Field & Induced Charge

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In summary, the dielectric constant of the dielectric is 2.07692. If Q=8nC, the area of the plates would be 5.45E-25 square meters.
  • #1
at3rg0
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Homework Statement



Two parallel plates have charges Q and -Q. When the space between the plates is devoid of matter, the electric field is 2.7E5 V/m. When the space is filled with a certain dielectric, the field is reduced to 1.3E5 V/m.

What is the dielectric constant of the dielectric?
I got this answer as 2.07692, which was correct.

If Q = 8 nC, what is the area of the plates?
I tried plugging the numbers into the formulas below, but I'm not getting the right answer...Am I missing a piece of relevant information?

What is the total induced charge on either face of the dielectric?
This will be easier once I figure out the second question.

Homework Equations


E=E(not)/kappa
V=Ed
C=Q/V
C=epsilon(not)*kappa*Area/distance

The Attempt at a Solution


I got the first answer, and the second I could not get.
 
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  • #2
at3rg0 said:

The Attempt at a Solution


I got the first answer, and the second I could not get.

It is more helpful to someone wishing to assist you if you show your calculation so they can see what you set up and what values you got. Thanks.

P.S. Are you given a plate separation distance?
 
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  • #3
Using the formulas, I got that

Area = Q*Epsilon(not)/Electric Field

So Area = (8E-9)(8.85E-12)/1.3E5 = 5.45E-25 m^2 (so 5.45E-23cm^2) - I need to give the answer in cm^2.
 
  • #4
at3rg0 said:
Using the formulas, I got that

Area = Q*Epsilon(not)/Electric Field

So Area = (8E-9)(8.85E-12)/1.3E5 = 5.45E-25 m^2 (so 5.45E-23cm^2) - I need to give the answer in cm^2.

The units on the right hand side are presently (C)·(C^2/N·m^2)/(N/C), so I don't think this is going to give you an area.

OK, you don't need a plate separation, but I suggest you review how you rearranged your equations to get Area...
 
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  • #5
Going by units alone...

Shouldn't Q/(Epsilon0 * E) give m^2?

All right, I tried rearranging again.

A = C/(Epsilon0*E*kappa), which gives me .003762. Where am I making the mistake in formula manipulation?

I used C=Q/V, where V=Ed, and C=kappa*epsilon0*Area/d
 
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  • #6
at3rg0 said:
Going by units alone...

Shouldn't Q/(Epsilon0 * E) give m^2?

All right, I tried rearranging again.

A = C/(Epsilon0*E*kappa), which gives me .003762. Where am I making the mistake in formula manipulation?

I used C=Q/V, where V=Ed, and C=kappa*epsilon0*Area/d

This is fine now. You set

C = (kappa·eps0·A)/d = Q/V , so

A = (Q · d)/(kappa · eps0 · V) , but E = V / d , so

A = Q / (kappa · eps0 · E).

The units are C / [ {(C^2)/N·(m^2)} · {N/C} ] = C / [C/(m^2)] = m^2 .

I also get your value, but it looks small because it's in m^2, so A = 33.6 cm^2. (What are you using for epsilon_0 ?)
 
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FAQ: Dielectric Problem: Q&A to Calculate Electric Field & Induced Charge

1. What is a dielectric material?

A dielectric material is an insulating material that does not easily conduct electricity. It is often used to separate two conductors in a capacitor, and its presence affects the electric field and charge distribution in the system.

2. How does a dielectric affect the electric field?

A dielectric material has a polarizing effect on the electric field, causing it to weaken or "screen" the electric field created by the charges on the conductors. This results in a decrease in the electric field strength and a redistribution of charge on the conductors.

3. What is the relationship between the electric field and induced charge in a dielectric?

The electric field and induced charge are directly related in a dielectric material. As the electric field weakens due to the presence of the dielectric, the induced charge on the surface of the dielectric increases. This is known as polarization.

4. How do you calculate the electric field in a dielectric?

The electric field in a dielectric can be calculated using the formula E = E0/κ, where E0 is the electric field in a vacuum and κ is the dielectric constant of the material. The dielectric constant is a measure of how much the material weakens the electric field.

5. What is the significance of dielectric constant in the calculation of electric field and charge?

The dielectric constant is an important factor in the calculation of electric field and charge distribution in a dielectric material. It determines the degree to which the electric field is weakened and the amount of induced charge on the surface of the dielectric. Different materials have different dielectric constants, which can greatly affect the behavior of the electric field in a system.

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