A parallel-plate air capacitor is made from two plates 0.200 m square, spaced 0.900 cm apart. It is connected to a 140 V battery. A. What is the capacitance?
B. What is the charge on each plate?
C. What is the electric field between the plates?
D. What is the energy stored in the capacitor?
E. If the battery is disconnected and then the plates are pulled apart to a separation of 1.40 cm , what are the answers to parts A, B, C, and D?
Enter your answer as four numbers corresponding to C, Q, E, U. Please enter the answer in the given order and in the same units as in parts A, B, C, and D.

See detailed solution below

Explanation:

a) From C= εoεrA/d

Where;

C= capacitance of the capacitor

εo= permittivity of free space

εr= relative permittivity

A= cross sectional area

d= distance between the plates

Since the relative permittivity of air=1 and permittivity of free space = 8.85 × 10^−12 Fm−1

Then;

C= 8.85 × 10^−12 Fm−1 × 0.2m^2/0.009 m

C= 196.67 × 10^-12 F or 1.967 ×10^-10 F

b) Q= CV = 1.967 ×10^-10 F × 140 V = 2.75 × 10^-8 C

c) E= V/d = 140 V/0.009m = 15.56 Vm-1

d) W= 1/2 CV^2 = 1/2 × 1.967 ×10^-10 F × (140)^2 =1.93×10^-6J

Part II

When the distance is now 0.014 m

a) C= 8.85 × 10^−12 Fm−1 × 0.2m^2/0.014 m = 1.26×10^-10 F

b) W= 1/2 Q^2/C = 1/2 × ( 2.75 × 10^-8 C)^2 / 1.26×10^-10= 3×10^-6 J

Note that the voltage changes when the distance is changed but the charge remains the same

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