Two circular metal discs each of 1.80 cm. diameter are spaced 0.55 mm apart to c

Question

Two circular metal discs each of 1.80 cm. diameter are spaced 0.55 mm apart to create a capacitor. Please do all the parts with shown equations

(1)What is the capacitance of this capacitor? If this capacitor is connected in series to a 25V battery, (b)What magnitude charge develops on each plate of the capacitor? While the battery remains connected, the plates are pulled apart until the separation between the plates in 1.20mm. (HINT: Think about whether it will remain constant in this scenario) (c)What is the capacitance now? (d) What potential difference is between the plates? (e)What charge develops on each plate of the capacitor?

2) Now the battery is disconnected and then the plates magically shrink from 1.80cm to 1.00cm.a)what is the capacitance now? b) what potential difference develops between the plates? c) what charge develops on each plate?d) what is the magnitude of E field?e) how much energy is stored in the capacitor?

Explanation / Answer

cross sectional area of the plate, A = pi*d^2/4

= pi*0.018^2/4

= 2.54*10^-4 m^2

distance between plates, d = 0.55*10^-3 m

1)
a)
C = A*epsilon/d

= 2.54*10^-4*8.854*10^-12/(0.55*10^-3)

= 4.09*10^-12 F

b)
Q = C*V

= 4.09*10^-12*25

= 1.02*10^-10 C

c)
C' = A*epsilon/d'

= 2.54*10^-4*8.854*10^-12/(1.2*10^-3)

= 1.874*10^-12 F

d) v' = v = 25 volts

2)

cross sectional area of the plate, A = pi*d^2/4

= pi*0.01^2/4

= 7.85*10^-5 m^2

C' = A*epsilon/d'

= 7.85*10^-5*8.854*10^-12/(1.2*10^-3)

= 5.8*10^-13 F

b) V' = V*C/C'

= 25*1.874*10^-12/(5.8*10^-13)

= 80.8 volts

c) Q' = C'*V'

= 5.8*10^-13*80.8

= 4.68*10^-11 C

d) E = v'/d

= 80.8/0.0.012

= 67333 N/c

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