You make a capacitor by cutting the 15.5-cm-diameter bottoms out of two aluminum

Question

You make a capacitor by cutting the 15.5-cm-diameter bottoms out of two aluminum pie plates, separating them by 3.40 mm , and connecting them across a 6.00 V battery.

Part A

What’s the capacitance of your capacitor?

Express your answer to three significant figures with the appropriate units.

Part B

If you disconnect the battery and separate the plates to a distance of 3.50 cm without discharging them, what will be the potential difference between them?

Express your answer to three significant figures with the appropriate units.

Explanation / Answer

part A) capacitance C = AEo/d

where Eo = 8.85*10^-12 F/m

A = pi*D^2/4 = pi*(0.155)^2/4 = 0.0188 m^2

d= 3.40 mm = 3.4*10^-3 m

=> C = 0.0188*8.85*10^-12/(3.4*10^-3) = 0.048*10^-9 F

part B) V = 6.0 v

charge on capacitor Q = CV = 0.048*10^-9*6 = 0.288*10^-9 Coulomb

if we disconnect the battery, Q remain the same , d= 3.50 cm = 0.035 m

potential difference V = Q/Cnew

=> Cnew = AEo/d = 0.0188*8.85*10^-12/0.035 = 4.753*10^-12 F

=> V = 0.288*10^-9/(4.753*10^-12) = 0.0605*10^3 volt

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