"In class, Dr. Brown slid a textbook between the plates of a parallel-plate capacitor after she had charged the capacitor and taken it out of the circuit with the battery. When she put the textbook between the plates, what happened to (a) the capacitance, (b) the charge on the plates, (c) the electric field E between the plates, and (d) the potential difference between the plates of the capacitor? How would your answers change if she had left the parallel-plate capacitor in the circuit with the battery while she inserted the textbook?"
C = [tex]\epsilon[/tex]A/d
V = Ed
Q = VC
The Attempt at a Solution
(a): From what I understand, just about any material has a greater permittivity than air. Since capacitance equals permittivity times area over distance, an increase in permittivity increases capacitance.
(b): Since the capacitor is not in a circuit with the battery, charge must be conserved, so the total charge on the capacitor must remain constant.
(c) and (d): Q=VC, C increases and Q is constant, so V must decrease. V = Ed and d is constant, so E must decrease as well.
In circuit with the battery: Instead of charge remaining constant, voltage remains constant (the voltage of the battery). Thus capacitance increases for the same reason as in the other case. Since V = Q/C, V is constant and C increases, Q must increase. Since V = Ed and V is constant, E must be constant as well.
Also, let's say that instead of putting a book between the plates, she pulled the plates apart. If the capacitor was not in circuit with a battery, the capacitance would decrease, charge would be constant, electric field would decrease, and voltage would increase. If it was in circuit with a battery, capacitance would decrease, charge would decrease, electric field would decrease, and voltage would be constant.
Is all of this right? Thanks a lot.