Electric charge of capacitor

[chmate]

Homework Statement

Dielectric medium inside capacitor consists of two types (check the pic). Capacitor's plates are connected to a source of constant tension U. Distance between the plates is d and their area is S. Find the electric charge of capacitor's plates.

Data given: U=500 V, d=2cm, S=80cm^2, εr1=2.5, εr2=1.0.
(ε1=εr1*ε0 ^ ε2 = εr2*ε0)

The Attempt at a Solution

By applying Gauss's Law for a surface around the capacitor's plate, I manage to get the electric field intensity which is $E=\frac{Q}{ε0εr1S1+ε0εr2S2}$

$U=\frac{Qd}{ε0εr1S1+ε0εr2S2}=500$

which implies

$Q=\frac{500(ε0εr1S1+ε0εr2S2)}{d}$

I don't know how to continue after this because I don't know S1 and S2 separately.

Help me!

 

[gneill]

If there's a variable left unspecified then your result will be an expression in that variable. Suppose you let r specify the division of the plate areas so that S1 = r*S and S2 = (1-r)*S.

 

[chmate]

Yes gneill, but still that doesn't solve the problem. Does this mean the book didn't give enough data?

 

[gneill]

Yes gneill, but still that doesn't solve the problem. Does this mean the book didn't give enough data?

It would seem that there are two possibilities:
1. The book mistakenly failed to provide enough data to solve the problem numerically.
2. The book intentionally did not provide enough data to solve the problem numerically and expects an expression for an answer.

A third but unlikely possibility is that the missing data doesn't matter if its variable somehow cancels out of the problem.

If this is a homework problem to be handed in, make note of the fact that there is not enough information for a complete numerical solution and provide a simplified expression for a result.

 

[Nickie]

It seems like you are on the right track in using Gauss's Law to solve for the electric field intensity. However, in order to find the electric charge of the capacitor's plates, you will also need to consider the relationship between electric field and electric charge.

We know that the electric field is defined as the force per unit charge, so we can write the equation E=\frac{F}{q}. In this case, the electric field is created by the electric charge on the plates, so we can rewrite the equation as E=\frac{Q}{ε0εr1S1+ε0εr2S2}.

Now, we can combine this equation with the one you have already found for the electric field intensity and solve for Q. This will give us the total electric charge on the plates.

Q=E(ε0εr1S1+ε0εr2S2)

Substituting in the values given in the problem, we get:

Q=\frac{500(ε0εr1S1+ε0εr2S2)}{d}(ε0εr1S1+ε0εr2S2)

Finally, we can solve for Q using the given values for d, S1, S2, εr1, and εr2. This will give us the electric charge of the capacitor's plates.