Help with capacitance question?

There are several ways to go about this, but as is usual in physics, the work-energy method is simplest. The energy stored in a capacitor with capacitance C, voltage V, and charge Q can be alternatively written as

U = 1/2 CV^2 = 1/2 VQ

Solving for Q,

Q = CV.

Note that this is the definition of capacitance itself:

C = V/Q

that is, how much electric potential V is established by an amount Q of electric charge. Just in case you don't remember, the capacitance of a parallel plate capacitor is

C = e_0*A/d

where e_0 is the electrical permittivity of vacuum, A is the capacitor area, and d is the distance between the plates.

(I'll let you do the math).

EDIT

------

Response to the poster below:

Of course the capacitance isn't given, but the geometry of the problem is completely specified. Since capacitance is a function only of geometry, I don't understand why you can't use Q = CV.

I quite understand your difficulty. You cannot use the formula C = Q / V to solve for Q because, although you know V, the capacitance C itself is not given.

Now, in solving part (a), I know you used the given electric field magnitude of 9.0e+05 N/C and the separation distance of 1.08 mm between the plates. Since you haven't use the given area of the plates, then, it must somehow be involved in solving for the charge Q. So, try to recall a general formula in electrostatics that involves the charge and the area. If you still cannot remember any such useful relation, try remembering a general property of charges. Don't they always produce an electric field. Now, if your unknown charge Q is somehow related to the given area A, the the electric field E produced by that charge must also be somehow related to the given area A. As the next step, try recalling a general formula in electrostatics that involves the electric field E and an area A. Doesn't that ring a bell concerning the concept of the electric flux or the flux of a vector field. Well, if you learned that earlier, then, you will recall that the flux of a vector field measures the amount of that vector passing or crossing through the area. In electrostatics, the flux of the electric field is given by Gauss's law

flux(E) = (epsilon zero) *surface integral [ E(dA)cos(theta) ] = total charge Q

If you take a very thin Gaussian surface that just encloses one plate, then you have your unknown charge Q on the right hand side of the last equation.

In the middle of the last equation, theta is the angle between the electric field vector and the unit normal vector perpendicular to the area element dA on the plate. Since E is also perpendicular to the area, then E and the unit normal are in the same direction making theta = 0 so that cos(theta) = 1.

In adition, the electric field vector between the plates is constant, so that you can take E out of the integral in Gauss's law. That leaves us only dA inside the surface integral and the surface integral of dA is just the whole area of the plate. We therefore get the following result from Gauss's law:

(epsilon zero) *(EA) = Q

Substituting the values

epsilon zero = 8.854e -12 C^2 / (N*m^2)

E = 9.00e+05 N/C

A = (0.92 cm^2)(1.0e - 02 m/cm)^2

you get your unknown charge Q.