Capacitance

Note by Talulah Riley
2 weeks, 2 days ago

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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 τ=RC \tau = R C increases

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

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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 - 2 weeks, 2 days ago

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

Talulah Riley - 2 weeks, 2 days ago

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@Steven Chase @Karan Chatrath Thanks in advance

Talulah Riley - 2 weeks, 2 days ago

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@Steven Chase I have uploaded a new problem above in this note only. Have a look. Thanks in advance

Talulah Riley - 2 weeks, 2 days ago

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With just S1 closed, there are 4 sources and 4 capacitors in series. So initially, each capacitor has voltage E. After closing S2, there are now two separate circuits, each with 2 sources and 2 capacitors in series. Since the ratio of sources to capacitors in each circuit is the same as in the initial case, the capacitor voltages remain the same, and no heat is dissipated.

Steven Chase - 2 weeks, 2 days ago

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@Steven Chase in have doubt . Let me share my rough work .

Talulah Riley - 2 weeks, 2 days ago

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@Steven Chase I have attached my rough work above. I am having great confusion. If we see clearly the potential difference across first capacitor is not E?

Talulah Riley - 2 weeks, 2 days ago

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@Talulah Riley The non-polarity side of a source is not at zero volts. It is at a potential E volts less than the polarity side. The ground points are all at zero volts.

Steven Chase - 2 weeks, 2 days ago

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@Steven Chase @Steven Chase But why E volts less. A battery ensures that it should hai potential differnce of E accross it. So why my work is incorrect?

Talulah Riley - 2 weeks, 2 days ago

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@Talulah Riley The potential at the polarity side of the battery is E volts greater than the potential at the non-polarity side. That does not mean that the voltage at the non-polarity side is "zero". In general, you can choose any single place in the circuit you want to be zero, and then you must be consistent from then on. All the ground (earth) points are at the same potential. By convention, I am saying that those points are at zero potential.

Steven Chase - 2 weeks, 2 days ago

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@Steven Chase @Steven Chase Ohkay i have again added my rough work. Is it correct now.?

Talulah Riley - 2 weeks, 2 days ago

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@Talulah Riley Yes, and x=E x = - E

Steven Chase - 2 weeks, 2 days ago

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@Steven Chase @Steven Chase how do you know x=Ex=-E????

Talulah Riley - 2 weeks, 2 days ago

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@Talulah Riley All the capacitor voltages are the same, since they are in series

Steven Chase - 2 weeks, 2 days ago

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@Steven Chase @Steven Chase Thanks for clearing doubts. I have added a new figure with the same problem.
Now, I am. Very interested to see how you will approach the problem.

Talulah Riley - 2 weeks, 2 days ago

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