A parallel-plate air capacitor is made by using two plates 15 cm square, spaced

Question

A parallel-plate air capacitor is made by using two plates 15 cm square, spaced 3.6 mm apart. It is connected to a 12-V battery. What is the capacitance? What is the charge on each plate? What is the electric field between the plates? What is the energy stored in the capacitor? The battery is disconnected and then the plates are pulled apart to a separation of 7.4 mm. What is the capacitance in this case? What is the charge on each plate in this case? What is the electric field between the plates in this case? What is the energy stored in the capacitor?

Explanation / Answer

(A) The capacitance

C = Eo*A/d

C = (8.85*10^-12)*(0.15)*(0.15)/(3.6*10^-3)

C = 5.53 * 10^-11 F

(B) The charge on each plate

Q = C*V

Q = (5.53*10^-11)*(12)

Q = 46*10^-9 C which is 46 nC

(C) The electric field between the plates

V = E*d

12 = (E)*(3.6 * 10^-3)

E = 3333.34 V/m

(D) The energy stored in the capacitor

E = 0.5*Q*V

E = (0.5)*(46*10^-9)*(12)

E = 2.76 * 10^-7 J

For the new distance,

(a) C = Eo*A/d

C = (8.85 * 10^-12)*(0.15)(0.15)/(7.4*10^-3)

C = 2.69*10^-7 F

(b)

Q can not change, so it stays constant at 46*10^-9 C which is 46 nC

(c)

V = E*d

12 = (E)*(7.4*10^-3)

E = 1621.62 V/m

(d)

Since Q and V are unchanged, E remains unchanged at 2.76 * 10^-7 J

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