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Maurinko [17]
3 years ago
13

The parallel plates in a capacitor, with a plate area of 7.10 cm2 and an air-filled separation of 2.20 mm, are charged by a 4.80

V battery. They are then disconnected from the battery and pulled apart (without discharge) to a separation of 6.50 mm. Neglecting fringing, find (a) the potential difference between the plates, (b) the initial stored energy, (c) the final stored energy, and (d) the work required to separate the plates.
Physics
1 answer:
Lostsunrise [7]3 years ago
6 0

Answer:

a)14.17V

b)32.8 x 10^-^1^2J

c)96.9x 10^-^1^2J

d) -64x 10^-^1^2J

Explanation:

Given:

Area 'A'=7.10cm² =>7.1 x 10^-^4m²

voltage 'V_o'=4.8 volt

d_o = 2.20mm => 2.2 x 10^-^3m

d_1 = 6.50mm => 6.5 x 10^-^3m

a) Capacitance C_o before push is given by:

C_o = εA/d_o =>\frac{(8.85*10^-^1^2)(7.1*10^-^4)}{2.2*10^-^3}

C_o =   2.85 x 10^-^1^2 F

q_o=C_oV_o=> 2.85 x 10^-^1^2 x 4.8

q_o=1.37 x 10^-^1^1 C

Capacitance C_1 after push is given by:

C_1 = εA/d_1 =>\frac{(8.85*10^-^1^2)(7.1*10^-^4)}{6.5*10^-^3}

C_1 =   9.66 x 10^-^1^3F

q_o=q_1

q_1=C_1V_1

Therefore, the potential difference between the plates

V_1 = 1.37 x 10^-^1^1 / 9.66 x 10^-^1^3 =>14.17V

b) U_i=\frac{1}{2}C_oV_o^2 => \frac{1}{2} (2.85*10^-^1^2)(4.8^2)

U_i= 32.8 x 10^-^1^2J

c)U_f=\frac{1}{2}C_1V_1^2 => \frac{1}{2} (9.66*10^-^1^3)(14.17^2)

U_f= 96.9x 10^-^1^2J

d) the work required to separate the plates is given by:

workdone=  U_i-U_f=> 32.8 x 10^-^1^2J- 96.9x 10^-^1^2J

W≈ -64x 10^-^1^2J

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