The energy stored in a capacitor (plate separation of 0.29 mm, square plates wit

Question

The energy stored in a capacitor (plate separation of 0.29 mm, square plates with sides 7.50 cm and kappa = 8.7), after being totally charged and disconnected from the voltage source, is used to melt a 6.7 mg sample of lead. To what voltage must the capacitor be charged, assuming the lead sample is initially at 18 degree C? If the plate separation is increased by 33%, how much lead, in mg, can be melted with the increase in the plate separation assuming the lead is at its initial temperature of 18 degree C?

Explanation / Answer

d = 0.29 * 10^-3 m
A = (7.5 * 7.5) * 10^-4 = 56.25 * 10^-4 m^2
k = 8.7

C = k*A*eo/d
C = (8.7 * 56.25 * 10^-4 * 8.85*10^-12)/(0.29 * 10^-3)
C = 1.49 * 10^-9 F

Energy stored in a capacitor is given by, U = 1/2*cv^2

Melting point of lead, = 327.5 °C
Energy needed to melt Lead, = m*Cl*t + m*Ll
Energy needed to melt Lead, = 6.5 * 10^-3 * 0.128 * (327.5 - 18) + 6.5*10^-3 * 22.4 = 0.403 J

Equating the two energy,
1/2*cv^2 = 0.403
1/2 * 1.49 * 10^-9 * v^2 = 0.403
v = 23258 volt

(b)
Now As the plate seperation is increased by 33% , New seperation = d + 0.33*d = 1.33 d
New Capacitance, C = k*A*eo/ 1.33d
C = (1.49 * 10^-9)/1.33 F
C = 1.12 * 10^-9 F

1/2 * Cv^2 = m*Cl*t + m*Ll
1/2 * 1.12 * 10^-9 * 23258^2 = m* 0.128 * (327.5 - 18) + m * 22.4
m = 4.9 mg
Mass of lead melted, m = 4.9 mg

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