Capacitance

#ElectricityAndMagnetism

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Comments

There are two possibilities:

1) The capacitor charge stays the same
2) The capacitor energy stays the same

It is not entirely clear to me which possibility is more valid. I will assume that the first is true. The consequences are:

1) The capacitor voltage decreases, so the current in the circuit decreases
2) The energy stored in the capacitor decreases
3) The capacitance increases, so the time constant $\tau = R C$ increases

Answer $D$ is true no matter which assumption we initially make. But the assumption of constant charge makes $D$ the only correct answer. See the code below for some trials and results.

import math

# General principles

# Q = C*V
# E = 0.5*C*(V**2.0)
# C is proportional to epsilon (dielectric strength)
# dielectric has permittivity greater than that of air

# Assumptions for when dielectric inserted:
# 1) Charge stays the same
# 2) Capacitance increases  

###########################################

# Case 1

Q1 = 2.0
C1 = 3.0

V1 = Q1/C1
E1 = 0.5*C1*(V1**2.0)

print V1
print E1
print ""
print ""

###########################################

# Case 2

Q2 = 2.0     # same as in Case 1
C2 = 5.0     # greater than in Case 1

V2 = Q2/C2
E2 = 0.5*C2*(V2**2.0)

print V2
print E2

###########################################

# Results

#>>> 
#0.666666666667
#0.666666666667


#0.4
#0.4
#>>>

Steven Chase - 8 months, 1 week ago

@Steven Chase yes D is the correct answer. Thanks for the solution.