Question Relating Electric Field and Capacitance

[rabcdred]

Homework Statement

A parallel-plate capacitor has a plate area of 0.2 m2 and a plate separation of 0.1 mm. To obtain an electric field of 2.0 × 10^6 V/m between the plates, the magnitude of the charge on each plate should be

Homework Equations

Q=epsilon(E)(A)

The Attempt at a Solution

I keep getting half of the correct answer. Does anyone know why I am off a factor of 2? Cheers.

 

[Antiphon]

Q=CV
C=epsilon*Area/separation

 

[ehild]

I keep getting half of the correct answer. Does anyone know why I am off a factor of 2?

No, without seeing your solution.

ehild

 

[binbagsss]

My guess would be that you are only considering the electric field caused by one of the plates. ?

 

[SammyS]

Without seeing your solution, my guess is that the correct answer is twice what your answer is.

 

[rabcdred]

Sorry about that.
Q=CV, C=(epsilon)A/d, V=Ed/ By substituting these last 2 equations into the first, the result is Q=(epsilon)A(E)-->8.85e-12(0.2)(2e6)= 3.54e-6. However, this is half of the correct answer. What am I doing wrong?

 

[ehild]

That is correct with the given data. But it is possible that the volume between the plates was filled by a dielectric, with dielectric constant of 2. Are you sure that the dielectric constant was not given?

ehild

 

[rabcdred]

Yes. The question I gave you was a copy paste from the document.

 

[ehild]

I have no more idea. It happens quite often that wrong solutions are given. ehild