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

In summary, the problem involves calculating the total energy stored in a parallel plate capacitor with dielectric material composed of two materials with different dielectric constants. The capacitor is pulled a distance x toward the positive x-axis and a potential difference V is applied between the plates. The resulting capacitance is a combination of a series of capacitances and a parallel capacitance. The formula for total energy stored is U=0.5C*V^2, where C is the resultant capacitance. The problem can be solved using the series and parallel laws of capacitance and substituting the appropriate values for the dielectric constants and dimensions.
  • #1
blueyellow

Homework Statement



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]

Homework Equations



U=0.5C*V^2

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
 
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  • #2
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
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
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
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]
 
  • #6
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:
  • #7
cupid.callin

yes, that is the equation
 
  • #8
cupid.callin:
thank you by the way - I'm sure that made what I meant a whole lot clearer
 
  • #9
bjd40@hotmail.com said:
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.
 
  • #10
yes, the three capacitors are in parallel
 
  • #11
what am I supposed to substitute in for V?
 
  • #12
blueyellow said:
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.
 

1. How do you calculate the total energy stored in a capacitor?

The total energy stored in a capacitor can be calculated using the formula E = 1/2 * C * V^2, where E is the total energy, C is the capacitance, and V is the voltage across the capacitor.

2. What is the relationship between the energy stored in a capacitor and its capacitance?

The energy stored in a capacitor is directly proportional to its capacitance. This means that as the capacitance increases, the energy stored also increases, and vice versa.

3. How does the voltage affect the energy stored in a capacitor?

The energy stored in a capacitor is directly proportional to the square of the voltage across it. This means that as the voltage increases, the energy stored also increases, and vice versa.

4. Can the energy stored in a capacitor vary with distance?

No, the energy stored in a capacitor is not affected by the distance between the plates. It is only affected by the capacitance and voltage.

5. What is the unit of measurement for energy stored in a capacitor?

The unit of measurement for energy stored in a capacitor is joules (J). It can also be expressed in other units such as electron-volts (eV) or calories (cal).

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