1-A parallel-plate air capacitor is made by using two plates 17 cm square, space

Question

1-A parallel-plate air capacitor is made by using two plates 17 cm square, spaced 3.7 mm apart. It is connected to a 12-Vbattery. If the battery remains conneated to the battery and the plates are pulled apart to a separation of 7.7 mm

a) What is the new capacitance?

b)What is the new charge on each plate?

c)What is the new electric field between the plates?

d)What is the new energy stored in the capacitor?

2- A 0.400-m-long cylindrical capacitor consists of a solid conducting core with a radius of 1.10 mm and an outer hollow conducting tube with an inner radius of 2.30 mm . The two conductors are separated by air and charged to a potential difference of 6.50 V .

a)Calculate the charge per length for the capacitor.

b)Calculate the total charge on the capacitor.

c)Calculate the capacitance.

d)Calculate the energy stored in the capacitor when fully charged.

Explanation / Answer

1-A:

area A = 17 sq.cm

distance d = 3.7 mm

Capcitance C = 8.854 x 10-12 x 17 x 10-4 /(3.7 x 10-3) -express all units in mts.

                        = 557 x 10-19 F

Wehn the plates are pulled apart to 7.7mm the new capcitance

C= (557 x 3.7 / 7.7) x 10-19 = 268 x 10-19 F

Voltage V = 12 V

Charge Q = C x V = 12 x 268 x 10-19 = 3216 x 10-19 C

Electric filed between the plates = V/d = 12/(7.7 x 10-3) = 1.56 x 103 V/m

Energy stored in the Capacitor = (1/2)QV = 0.5 x 12 x 3216 x 10-19 = 19296 x 10-19 J

Part 2-A Cylindrical capcitor

PD = 6.5 V, inner radius r = 1.1 mm, outer radius R = 2.3 mm length 0.4 m

Capcitance per unit length = 2x pi x e0 /(ln(R/r) = 2 x 3.14 x 8.854 x 10-12 /ln(2.3/1/1) = 75.38 x 10-12 F

Charge per unit length = 75.38 x 10-12 x 6.5 = 489.97 x 10-12 C

Total charge C = 489 x 10-12 x0.4 = 198.98 x 10-12 C

Total Capacitance C = 75.38 x 0.4 x 10-12 =30.152 x 10-12

Energy stored = 0.5 QV = 0.5 x 198.98 x 6.5 x 10-12 = 646.68 x10-12 J

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