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? ΔV_new= _______________ ΔV_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? σ_new= _____________ σ_old
(f) Consider how E is related to the charge density, by what fraction does the E field change? E_new= _______________ E_old.

Answer 1

This problem is from topic capacitor.