Question

A parallel-plate capacitor is charged and then disconnected from a battery.
Then, the plate separation is decreased by a factor of 4, d_new=1/4 d_old,
please type in a number into the box to determine how everything changes.
(If something doesn't change, please type in 1 to answer X_new= 1 X_old.)
(a) By what fraction does the capacitance change? C_new=  C_old;
(b) By what fraction does the amount of charge change? Q_new=  Q_old (When the battery is disconnected, the charges have no place to go to and will stay where they were).
(c) By what fraction does the voltage difference between the two plates change? ?V_new=  ?V_old
(When the battery is disconnected,voltage difference may no longer be kept to be the same value any more.).
(d) By what fraction does the energy stored in the capacitor change? U_new=  U_old
Does it make sense? Let's check, considering the amount of charge and the area of the plate,
(e) by what fraction does the amount of charge density change? ?_new=  ?_old
(f) Consider how E is related to the charge density, by what fraction does the E field change? E_new=  E_old.
Considering how E, d and voltage difference are related, does it make sense now, by what fraction does the Voltage difference change?
Please explain why the energy stored in the capacitor changes? If it increased, where did the energy come from? If it decreased, where could the energy go? (The force between the two plates are attracting or repelling?. When the plate separation decreases, does the electric force do positive work or negative work?)

Answer 1

(a) capacitance of capacitor is given by C = Area/distance

When d' = d/4

Then C' = 4C

(B) when we disconnect the battery then charge remains constant on capacitor as there is no interaction with other matter.

(c) Voltage across capacitor is given by V = Q/C , Q is charge on capacitor.

V' = Q/4C = V/4

(d)potential energy stored in capacitor = 0.5CV²

U' = 0.5×4C×(V/4)² = U/4

(e) volume charge density will becomes 4 times but surface charge density on plate will remain same.

(f)Electric field is ?V/d

E' = (?V/4)/(d/4) = ?V/d = E will remain same