How Does Changing Plate Separation Affect Energy Stored in a Capacitor?

[eku_girl83]

I need some help with these problems:
1) A parallel-plate vacuum capacitor has 7.36 J of energy stored in it. The separation between the plates is 3.64 mm. If the separation is decreased to 1.40 mm, what is the energy stored if the following events occur?
a) the capacitor is disconnected from the potential source so the charge on the plates remains constant
b) the capacitor remains connected to the potential source so the potential difference between the plates remains constant
I have the equations U=.5CV^2=.5QV=(Q^2)/(2C)
.5CV^2/Ad
C=Epsilon_0(A)/d, V=Ed, u=.5Epsilon_0*E^2
How do I apply them to this problem?

2) A capacitor has parallel plates of area 20 square centimeters separated by 1.4 mm. The space between the plates is filled with polycarbonate (dielectric constant = 2.8, dielectric strength = 3*10^7)
I correctly calculted the permittivity (2.479E-11) and maximum permissible voltage (42000). I need help finding the surface-charge density on each plate and the induced surface charge density on the surface of the dielectric. I really don't know where to start with this, so any help would be appreciated!

Thanks for help with one or both questions,
eku_girl83

 

[kartiksg]

Energy of capacitor U = 1/2 CV^2. If voltage is constant and you bring the plates together, V stays put but C increases. So the energy increases.

U = 1/2 Q^2/C. If u disconnect the power supply, Q stays put, but C increases. So U decreases. I think u can plug in and calculate the actual values.

For the next problem, what voltage do u apply?
I just take it as V volts.
Ed = V
E = Efree/k (Efree is the free charge and k is the dielectric constant)

E = Efree - Einduced (Considering magnitudes only.. vectorially, they would add up).

=> Einduced = (1 - (1/k))Efree
=> sigma induced = (1-1/k) sigma free (sigma is surface charge density)

Now u can calculate sigma free easily using C = Q/V and then dividing Q by surface area. Using that, calculate sigma induced.

Hope it helped

Kartik

 

[M98Ranger]

1) To solve this problem, we can use the equation U = 0.5CV^2, where U is the energy stored in the capacitor, C is the capacitance, and V is the potential difference between the plates. We can also use the equation C = ε0A/d, where ε0 is the permittivity of free space, A is the area of the plates, and d is the separation between the plates.

a) If the capacitor is disconnected from the potential source, the charge on the plates remains constant. This means that the potential difference between the plates will also remain constant. Therefore, we can set the initial and final potential differences equal to each other and solve for the final energy stored.

Initial energy: U1 = 0.5C1V1^2 = 7.36 J
Final energy: U2 = 0.5C2V2^2

Since the charge remains constant, we can also set the initial and final capacitances equal to each other. So, C1 = C2 and we can solve for C2 using the equation C = ε0A/d.

Initial capacitance: C1 = ε0A/d1
Final capacitance: C2 = ε0A/d2

Substituting these values into the equation for final energy, we get:

U2 = 0.5C2V2^2 = 0.5(ε0A/d2)V2^2

Since the potential difference remains constant, we can set V1 = V2 and solve for V2 using the equation V = Ed.

Initial potential difference: V1 = Ed1
Final potential difference: V2 = Ed2

Substituting this into the equation for final energy, we get:

U2 = 0.5(ε0A/d2)(Ed2)^2

Now, we can substitute the given values for A, d, and U1 into this equation and solve for U2:

U2 = 0.5(8.85*10^-12 F/m)(0.02 m^2)/(0.0014 m) * (0.0014 m * (7.36 J / 0.02 m^2)^2

U2 = 0.000181 J

Therefore, the final energy stored in the capacitor is 0.000181 J.

b) If the capacitor remains connected to the