A 100-microCoulomb parallel-plate capacitor is charged with a 24V power supply, then is disconnected from the power supply.
a) The plates are moved 0.01 m apart to 0.05 m apart. What happens numerically, to the:
i. Capacitance of the device?
ii. Potential difference between the plates?
iii. Charge on the plates?
iv. Amount of energy stored in the capacitor?
b) A dielectric with constant k=4 is inserted into the gap (which is still 0.05m). What happens to the:
i. Capacitance of the device?
ii. Potential difference between the plates?
iii. Charge on the plates?
iv. Amount of energy stored in the capacitor?
v. When the dielectric is inserted, is the dielectric pulled in between the plates, or do you have to push it in there? Explain.
C = e A/d = 10^-4
Q = C V = e A V/D = 24*10^-4 =
U = (1/2) C V^2 = (1/2) Q V = 288*10^-4
a)increase d by factor of 5
i capacitance 1/5 of original
ii V = Q/C = 5 times original (you had to do work to pull those plates apart)
The charges stay on the plates so iii Q the same
iv U = (1/2) Q V but Q the same so five times the potential energy.
b is just more of the same. You did nothing to add or subtract Q so the charge is the same and work out from there.
if in iv the U goes up. you had to push in to raise the potential energy stored. If U goes down, the dielectric gets pulled in
Thanks again! and I will definitely check your arithmetic to make sure. You are seriously the best helper here! =]
a) To answer the questions, we need to understand the formulas and concepts related to capacitors:
1. Capacitance (C): The capacitance of a parallel-plate capacitor is given by the formula C = ε0 * (A / d), where ε0 is the permittivity of free space (approximately 8.85 x 10^-12 F/m), A is the area of the plates in square meters, and d is the distance between the plates in meters.
2. Potential Difference (V): The potential difference between the plates of a capacitor is equal to the voltage across the capacitor, which remains constant if the capacitor is disconnected from the power source. So, in this case, the potential difference remains the same at 24V.
3. Charge (Q): The charge on the plates of a capacitor is given by the formula Q = C * V, where Q is the charge in Coulombs, C is the capacitance in Farads, and V is the potential difference in Volts.
4. Energy Stored (U): The energy stored in a capacitor is given by the formula U = 0.5 * C * V^2, where U is the energy in Joules, C is the capacitance in Farads, and V is the potential difference in Volts.
Now let's answer the questions:
a) i. Capacitance (C): The capacitance of a parallel-plate capacitor is inversely proportional to the distance between the plates. Therefore, as the plates are moved from 0.01m apart to 0.05m apart, the capacitance decreases. To find the new capacitance value, we use the formula C = ε0 * (A / d), where ε0 is a constant for free space and A is the area of the plates. As the area is not given, we can assume it remains constant. So, the new capacitance can be calculated as C' = ε0 * (A / d'), where d' is the new distance. Plug in the values to get the new capacitance.
ii. Potential Difference (V): The potential difference between the plates of a capacitor remains constant when disconnected from the power supply. So, the potential difference remains 24V.
iii. Charge (Q): The charge on the plates of a capacitor is given by Q = C * V. As we know the capacitance and potential difference, we can calculate the charge.
iv. Amount of Energy Stored (U): The amount of energy stored in a capacitor is given by U = 0.5 * C * V^2. Using the new capacitance (obtained in part i) and the potential difference, calculate the amount of energy stored.
b) i. Capacitance (C): When a dielectric is inserted into the gap between the plates of a capacitor, the capacitance increases. The new capacitance can be calculated using the formula C' = k * C, where C is the initial capacitance and k is the dielectric constant of the material. In this case, k = 4, so multiply the initial capacitance by 4 to find the new capacitance.
ii. Potential Difference (V): The potential difference between the plates remains the same when a dielectric is inserted, so it remains 24V.
iii. Charge (Q): The charge on the plates also remains the same when a dielectric is inserted, so it is the same as in Part a.
iv. Amount of Energy Stored (U): The amount of energy stored in a capacitor with a dielectric is given by U' = 0.5 * C' * V^2. Use the new capacitance and the potential difference to calculate the new amount of energy stored.
v. When the dielectric is inserted, it is pulled in between the plates. The dielectric experiences an electric field created by the charged plates, attracting it towards the middle. Note that this attraction force may not be strong enough to pull the dielectric completely in between the plates if there are any physical obstructions preventing it. In most cases, the dielectric is easily inserted due to the attraction force.