Question

a variable capacitor consists of two plates with area 1.00 cm^2. The space between them is filled with material with dielectric constant 1.40. The plate separation can be adjusted by turning a finely threaded screw. Each revolution of the screw changes the separation by 0.010 cm. The initial plate separation is 0.040 cm.

A) A 12V battery is connected to the capacitor. What are the capacitance, voltage and charge of the capacitor?

B) With the 12V battery connected, the screw is tightened (bringing the plates closer) one half revolution. What are the capacitance, voltage and charge of the capacitance afterward?

C) The battery then is removed, leaving the charge on the capacitor. Then the screw is loosened one half revolution. What are the capacitance, voltage and charge of the capacitor now?

Answer 1

Capacitance of a parallel plate capacitor is given by

C = kAe0/d
k: dielectric constant

A) C = 1.4*0.0001*8.85*10^-12/0.0004
C = 3.0975*10^-12 Farad

Voltage across capacitor = 12 V (voltage of battery)

Charge, Q = CV = (3.0975*10^-12)*12 = 37.17*10^-12 Coulomb

B) One half revolution = 0.01/2 = 0.005 cm
New d = 0.04-0.005 = 0.035 cm

C = 1.4*0.0001*8.85*10^-12/0.00035

C = 3.54*10^-12 Farad

Voltage across capacitor = 12 V (voltage of battery)

Charge, Q = CV = (3.54*10^-12)*12 = 42.48*10^-12 Coulomb

C) Now, no battery hence charge will be conserved

Q = 42.48*10^-12 Coulomb

New d = 0.035+0.005 = 0.04 cm (loosened)

C = 1.4*0.0001*8.85*10^-12/0.0004

C = 3.0975*10^-12 Farad

V = Q/C = 42.48*10^-12 Coulomb/3.0975*10^-12 Farad = 13.714 Volts