Solve Capacitance Problem: Voltage, Charge & Dielectric Effects

[Xyius]

I do not know whether this is the correct section to post this. This is in my graduate level Electromagnetism class but the problem is sort of a "throw back" to the freshman level course. So I very much apologize if this is not the correct place!

This problem shows a parallel plate capacitor with area 1m^2 and a distance of 1mm between the plates and a voltage of 100V across it.. Basically the problem has many parts and is a sequential type problem.

a.)With the dielectric of air [itex]ε_r=1.0[/itex] find the capacitance. With the switch closed, how much charge is stored?

Seems easy enough, I just used the following formulas.

[itex]C=ε_0\frac{A}{d}=(8.84\times 10^{-12})\frac{1}{0.001}=8.84\times 10^{-9}F[/itex]
[itex]Q=CV=8.84\times 10^{-9}(100)=8.84\times 10^{-7}C[/itex]

b.)A dielectric is placed between the plates with a substance of a relative permitivity of 2.1, find the capacitance and the charge stored.

So I pretty much used the same two formulas except instead of [itex]ε_0[/itex] I used [itex]ε_0ε_r[/itex].

[itex]C=ε_0ε_r\frac{A}{d}=(8.84\times 10^{-12})(2.1)\frac{1}{0.001}=1.86\times 10^{-8}F[/itex]
[itex]Q=CV=1.86\times 10^{-8}(100)=2\times 10^{-6}C[/itex]

So, assuming I did these parts correctly, they weren't too bad. The following parts of the problem are what confuse me as I am not sure if I am able to show them mathematically or if it is enough to just explain.

c.)If someone were to open the switch to this capacitor, what happens to the voltage and charge between the plates? (Remember the material is still between the plates!)

So this is what I believe would happen. Since there is no more voltage supply, the backwards electric field from the dielectric would cause the electric field between the plates to become less. Since the electric field becomes less, the voltage would drop.

As for the charge, my gut instinct is to think that since the voltage is dropping, and the electric field is dropping, the charge should also drop as well.

d.)While the switch is open, the material between the plates is removed. What happens to the charge and the voltage?

Well since the material is gone, there is no back electric field being produced in the material and the electric field between the plates increases, this consequently causes an increase in the voltage and the charge as well.

What do you guys think? I do not feel confident in these answers as often times I noticed that relying on gut instincts (which I did in this type of problem) usually leads to the wrong type of thinking. Could I show these answers mathematically?

Thanks!

 

[phyzguy]

I think you've done a) and b) correctly, but you need to think more about c). One question to think about is, if your answer is correct, where did the charge that was stored on the capacitor go?

 

[Xyius]

I think you've done a) and b) correctly, but you need to think more about c). One question to think about is, if your answer is correct, where did the charge that was stored on the capacitor go?

Ah that is a very good point! I guess since the switch is open, there really isn't anywhere for it to go correct? That would mean that Q stays constant. Am I correct in saying the voltage would drop? That would mean in the equation Q=CV, the capacitance must go up to keep the equation balanced.

So for d, when the material is removed, the voltages goes up (for reasons explained in my first post) and the capacitance goes down to keep Q the same.

Any closer?? :\

 

[phyzguy]

Ah that is a very good point! I guess since the switch is open, there really isn't anywhere for it to go correct? That would mean that Q stays constant. Am I correct in saying the voltage would drop? That would mean in the equation Q=CV, the capacitance must go up to keep the equation balanced.
Any closer?? :\

Why should the voltage or capacitance change in part c)? Since C = ε A/d, did any of these values change?

 

[Xyius]

Why should the voltage or capacitance change in part c)? Since C = ε A/d, did any of these values change?

Hmm..So basically nothing changes when you disconnect the circuit? It was my understanding that a back electric field was produced in the dielectric due to the external electric field creating dipoles in the material. I assumed that since the electric field changed, the voltage would also change.

 

[phyzguy]

Hmm..So basically nothing changes when you disconnect the circuit?

Correct. Once the capacitor is charged up, it stays charged up.

 

[Xyius]

Okay that makes sense. But surely the capacitance would go down when the dielectric is removed correct?

 

[vela]

Yes, removing the dielectric causes the capacitance to decrease, but your answer to (d) above isn't correct.