Maximum charge on parallel-plate capacitor with field limit

[imationrouter03]

I'm currently in a physicsw/calII class and I'm stuck on this problem, hope u can help me out.

A parallel-plate capacitor has a capacitance of C_0 when there is air between the plates. The separation between the plates is x.

a)What is the maximum magnitude of charge that can be placed on each plate if the electric field in the region between the plates is not to exceed V?

For this question a, I've tried q=(V*epsilon*A)/x but the answer did not depend on the variables: A, and epsilon_0
I've also tried q=C_0*V but the answer involved the variable x. =(

b) A dielectric with a dielectric constant of K is inserted between the plates of the capacitor, completely filling the volume between the plates. Now what is the maximum magnitude of charge on each plate if the electric field between the plates is not to exceed V?

For this question i think the dielectric wouldn't change the charge value but it would affect the capacitor by a factor of K. I don't know where to go from here.. b.c "not to exceed V" i don't understand that statement.

thank you for your time and concern.. any feedback would be appreciated )

 

[harsh]

The voltage is the ratio of the charge and the capacitance. V = Q / C. Therefore, maximum charge should be independent of the distance between the plates.

When you add the dielectric to a capactior, you increase the capacitace by K times. The new capacitance would be K * C. If the voltage across has to be kept constant, and the capacitance increases, then the charge also has to increase by that value i.e. K. Therefore, you new charge value would be K times your old charge value.

- harsh

 

[quantumdude]

a)What is the maximum magnitude of charge that can be placed on each plate if the electric field in the region between the plates is not to exceed V?

For this question a, I've tried q=(V*epsilon*A)/x but the answer did not depend on the variables: A, and epsilon_0
I've also tried q=C_0*V but the answer involved the variable x. =(

Careful, if V in this problem corresponds to electric field (as opposed to electric potential), then this V does not mean the same thing as the V in the formula "q=CV". I think it's an incredibly bad idea to use V for the E field, so I won't. I'll use E.

So, if the max E field is E, then the max potential difference is V=Ex.

That should help with both parts. Give it a shot, and come back if you need to.

 

[imationrouter03]

well I've tried : k*E*x*C but that didn't seem to work.. it said that the answer doesn't involve the variable k nor C

i don't know what to do.. i don't even think i understand the problem

feedback please

 

[eJavier]

The field E between the plates is $$\frac{1}{\epsilon_0} \sigma$$ in SI units, that should give you a hint

 

[imationrouter03]

i used E=sigma/epsilon_0 and equated V=Ex with V=(sigma*x)/epsilon_0 and got the following solving for the charge

q=(V*A*epsilon)/(x)

which just happened to be my first wrong response...

The correct answer shouldn't involve variables: A, epsilon_0, k nor C.. what am i doing wrong?

feedback please