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

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

[itex]C \ = \ \epsilon_0 L \frac{(x + 0.6L)\epsilon_1 \ + \ \ (0.4 - x)\epsilon_2}{d} [/itex]
 
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:
cupid.callin

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

gneill

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no, the resultant c is of the form (c1*c2/c1+c2) + c0. just figure it from the series and parallel laws of capacitance.
Looks to me that all three capacitors should be in parallel.
 
yes, the three capacitors are in parallel
 
what am I supposed to substitute in for V?
 

gneill

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what am I supposed to substitute in for V?
:confused: 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.
 

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