PLEASE HELP!!! In the following problem, C= capacitance of the capacitor, Q = ma

Question

PLEASE HELP!!!
In the following problem, C= capacitance of the capacitor, Q = magnitude of charge on each plate, V = voltage difference between the capacitor plates, Enet = net electric field between the plates, and U = potential energy stored on capacitor.)

A gigantic parallel plate capacitor has two plates, each with an area of 7 m2, and a separation of 3 nm between them. You may assume a vacuum in between the plates before the dielectric is inserted.

a.) Calculate the capacitance of this capacitor.
Co = mF
(Note the 'C' value, considering how physically large the plates are.)

b.) The capacitor is now connected to a 12 V battery. When it is fully charged, calculate the amount of charge on each plate, the electric field between the plates (far from the edges), and the total energy it stores. Also, find the magnitude of the area charge density on each plate.
Qo = C
Uo = J
Eo = N/C
|?o| = C/m2

c.) If the capacitor was then disconnected from the battery, and then a dielectric with a constant of ? = 3 inserted between the plates, find the new values for all of the following (after the system has reached equilibrium.) You may express them as multiples of the initial values (from parts a-b).
C =

Explanation / Answer

a) C = e0 A/d = 8.85E-12*7/3.0E-9= 20.65 mF

b) Q = CV =20.65E-3*12= 0.248 C

U = 1/2 C V^2 = 0.5*20.65E-3*12^2= 1.49 J

E = V/d = 12/3.0E-9 = 4.0E9

sigma = Q/A = 0.248/7 =0.0354

c) now Q will be the same

C = 3 C0

Q = 1 Q0

since Q = CV

V = 1/3 V0

Enet = V/d = 1/3 E0

sigma = Q/A = 1*sigma0

U = Q^2/2C = Q0^2/(2*3C0) = 1/3 U0

d) now v stays the same so
C = 3 C0
Q = 1/3 Q0
V = V0
E = E0
sigma = 1/3 sigma0
U = 1/2 C V^2 = 3 U0

e) 4.0E9

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