1) To find the energy of the dielectric-filled capacitor (U_1), we use the formula:
U_1 = (1/2) * C_1 * V^2
Now, we know that capacitance C_1 depends on the dielectric constant k, the plate_area A and the plate separation d, and it is given by:
C_1 = k * epsilon_0 * (A/d)
Plugging in the values, we find:
C_1 = 5.00 * 8.85×10^(-12) C^2/N*m^2 * (30.0 x 10^(-4) m^2 / 10.0 x 10^(-3) m)
C_1 = 13.275 x 10^(-12) F.
Now, we use this value of C_1 to find the energy U_1:
U_1 = (1/2) * 13.275 x 10^(-12) F * (7.50 V)^2
U_1 = 2.983 × 10^(-7) J.
2) Now, the capacitor remains connected to the battery while the dielectric is pulled halfway out. The effective capacitance of the entire capacitor becomes the sum of the capacitance of the portion filled with the dielectric (C1') and the portion without the dielectric (C2').
C1' = (1/2) * k * epsilon_0 * (A/d) = (1/2) * C_1 = 6.6375 × 10^(-12) F
C2' = (1/2) * epsilon_0 * (A/d) = (1/5) * C_1 = 2.655 × 10^(-12) F
The total capacitance (C') of the half-filled capacitor is therefore:
C' = C1' + C2' = 6.6375 × 10^(-12) F + 2.655 × 10^(-12) F = 9.2925 × 10^(-12) F
Now, we can find the energy U_2 of the half-filled capacitor:
U_2 = (1/2) * C' * V^2
U_2 = (1/2) * 9.2925 × 10^(-12) F * (7.50 V)^2
U_2 = 2.084 × 10^(-7) J.
3) When the dielectric is removed completely and the capacitor is disconnected from the battery, the energy of the capacitor (U_3) will depend on the original capacitance of the capacitor without the dielectric (C0) and the final voltage across the capacitor (V_f).
U_3 = (1/2) * C0 * V_f^2
From the problem, we know that C0 = epsilon_0 * (A/d), which is equal to (1/5) * C_1.
We can find the final voltage across the capacitor (V_f) by using the fact that the stored charge in the capacitor remains constant. Therefore, for the same charge:
C0 * V_f = C' * V => V_f = (C'/C0) * V
Plugging in the values, we find:
V_f = (9.2925 × 10^(-12) F / 2.655 × 10^(-12) F) * 7.50 V
V_f = 3.75 V
Now, we use this to find the energy U_3:
U_3 = (1/2) * 2.655 × 10^(-12) F * (3.75 V)^2
U_3 = 9.481 × 10^(-8) J.
4) The work done by the external agent in removing the remaining portion of the dielectric from the disconnected capacitor can be found by calculating the difference between the initial and final energies of the capacitor (U_2 and U_3):
W = U_2 - U_3
W = 2.084 × 10^(-7) J - 9.481 × 10^(-8) J
W = 1.136 × 10^(-7) J.
Energy of a Capacitor in the Presence of a Dielectric
A dielectric-filled parallel-plate capacitor has plate area A = 30.0 cm^2, plate separation d = 10.0 mm and dielectric constant k = 5.00. The capacitor is connected to a battery that creates a constant voltage V = 7.50 V. Throughout the problem, use epsilon_0 = 8.85×10−12 C^2/N*m^2.
1)Find the energy U_1 of the dielectric-filled capacitor. (Find U_1)
2)The dielectric plate is now slowly pulled out of the capacitor, which remains connected to the battery. Find the energy U_2 of the capacitor at the moment when the capacitor is half-filled with the dielectric. (Find U_2)
3)The capacitor is now disconnected from the battery, and the dielectric plate is slowly removed the rest of the way out of the capacitor. Find the new energy of the capacitor, U_3
4)n the process of removing the remaining portion of the dielectric from the disconnected capacitor, how much work W is done by the external agent acting on the dielectric?
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