A parallel plate capacitor has an area of 5.00 cm 2 and a capacitance of 3.50 pF
ID: 1551552 • Letter: A
Question
A parallel plate capacitor has an area of 5.00 cm 2 and a capacitance of 3.50 pF. The capacitor is connected to a 12.0 V battery. After the capacitor is completely charged up, the battery is removed.
(A) What must be the separation of the plates?
(B) What is the charge density on the plates?
(C) What is the magnitude of the electric field between the plates?
(D) How much energy is stored in the electric field between the plates?
(E) What is the energy density in the space between the plates?
(F) If a sheet of Mylar is placed between the plates, how will the voltage across the plates and the charge density on the plates change?
A parallel plate capacitor has an area of 5.00 cm 2 and a capacitance of 3.50 pF. The capacitor is connected to a 12.0 V battery. After the capacitor is completely charged up, the battery is removed.
(A) What must be the separation of the plates?
(B) What is the charge density on the plates?
(C) What is the magnitude of the electric field between the plates?
(D) How much energy is stored in the electric field between the plates?
(E) What is the energy density in the space between the plates?
(F) If a sheet of Mylar is placed between the plates, how will the voltage across the plates and the charge density on the plates change?
Explanation / Answer
(A)
C= eo A/ d
d = eo A/C
= 8.85 * 10^-12 ( 5.0 * 10^-4)/3.5 * 10^-12
=1.26 mm
(b) charge density =Q/A = CV / A = 3.5 * 10^-12 ( 12)/5.0 * 10^-4
=8.4 * 10^-8 C/m^2
(c)
E= V/d = 12/1.26 * 10^-3 m = 9523.8 V/m
(d)
E = 1/2 CV^2 = 1/2 ( 3.5 * 10^-12) ( 12)^2 = 0.25 * 10^-9 J
(e)
energy density = E/A = 0.25 * 10^-9 J/5.0 * 10^-4 = 5.0* 10^-7 J/m^2
(f)
V' = V/k = 12 /3.1 = 3.87 V
charge density = CV' / A = 3.5 * 10^-12 ( 3.87)/5.0 * 10^-4
=2.7 * 10^-8 C/m^2