# Calculate the total energy stored in the capacitor as a function of x

1. Jul 16, 2011

### blueyellow

1. The problem statement, all variables and given/known data

C=epsilon(subscript0)L[[x+0.6L*epsilon(subscript1)+(0.4-x)epsilon(subscript2)]/d]

calculate the total energy stored in the capacitor, as a function of x [4 marks]

2. Relevant equations

U=0.5C*V^2

3. The attempt at a solution

I don't know what I am supposed to do here, if I just substitute C=... into the above formulae, it probably won't give me the answer they are looking for. Is there another equation I should use that is not the equation I mentioned? I do not know what Q is

2. Jul 16, 2011

### bjd40@hotmail.com

what do they mean by x, L and d here? they don't seem to be constants. what is the statement of the problem???

3. Jul 16, 2011

### blueyellow

here are some earlier parts of the question:
consider a parallel plate capacitor with square plates of side L and distance d<<L apart. The bottom plate lies on the x-y plane, and the distance d is parallel to z. A block of dielectric material with dimensions (L*L*d) can completely fill the space between the plates.

Let us consider instead the dielectric to be composed of two materials glued together, material 1 with dielectric constant epsilon1 and dimensions 0.6L*L*d(in x, y and z directions respectively) and material 2 with constant epsilon2 and dimensions 0.4L*L*d. The dielectric is free to move as a single block without friction along the x axis, parallel to the plates inside the capacitor, and it can also move outside the capacitor. Let us define as x the distance between the dielectric and the edge of the plate, along the x axis. A potential difference V is applied between the plates, and we can neglect the electric field outside the plates.

4. Jul 18, 2011

### bjd40@hotmail.com

sorry, i was offline a bit long.
the key to this problem is to calculate the capacitance of the parallel plate when the combined slab is pulled a distance x toward +X axis. if u draw the figure u will see that the combined capacitance can be calculated from the combination of a series of capacitances containing dielectric e1 and e2 and with that combination in parallel a capacitance containing dielectric of e0 . now calculate c1, c2 and c0 by using the formula c = e*area/distance of seperation. be careful while putting the expressions of area of each capacitor. find the combined capacitance and put the expression in U=05c*v2. a bit of algebra is needed. u can find U.

5. Jul 18, 2011

### cupid.callin

is this the eqn you have written:

$C \ = \ \epsilon_0 L \frac{(x + 0.6L)\epsilon_1 \ + \ \ (0.4 - x)\epsilon_2}{d}$

6. Jul 18, 2011

### bjd40@hotmail.com

no, the resultant c is of the form (c1*c2/c1+c2) + c0. just figure it from the series and parallel laws of capacitance.

Last edited: Jul 19, 2011
7. Aug 20, 2011

### blueyellow

cupid.callin

yes, that is the equation

8. Aug 20, 2011

### blueyellow

cupid.callin:
thank you by the way - I'm sure that made what I meant a whole lot clearer

9. Aug 20, 2011

### Staff: Mentor

Looks to me that all three capacitors should be in parallel.

10. Aug 20, 2011

### blueyellow

yes, the three capacitors are in parallel

11. Sep 3, 2011

### blueyellow

what am I supposed to substitute in for V?

12. Sep 3, 2011

### Staff: Mentor

V is just V. The problem statement says that a potential V is applied. Your result should be an expression that involves the symbol V along with the other parameters of the problem.