Homework Answers
a)
Voltage across the capacitor
V=Ed =(3*106)(2*10-3)
V=6000 Volts
Capacitance
C=eoA/d =(8.85*10-12)(2.5*10-3)/(2*10-3)
C=1.11*10-11F
Maximum charge
Qmax=CV =(1.11*10-11)*6000
Qmax=6.64*10-8C =7*10-8 C (approx)
b)
Voltage across the capacitor
V=Ed =(24*106)(2*10-3)
V=48000 Volts
Dielectric constant k=2.56
Capacitance
Cp=KC =2.84*10-11F
Maximum charge
Qmax=CV =(2.84*10-11)*48000
Qmax=1.36*10-6C =1.4*10-6 C (approx)
Add Answer to:
a= 7*10^-8
b = 1.4*10^-6
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