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.
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.
3 answers
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
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! =]
1 answer