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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.

importmath# 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 1Q1=2.0C1=3.0V1=Q1/C1E1=0.5*C1*(V1**2.0)printV1printE1print""print""############################################ Case 2Q2=2.0# same as in Case 1C2=5.0# greater than in Case 1V2=Q2/C2E2=0.5*C2*(V2**2.0)printV2printE2############################################ Results#>>> #0.666666666667#0.666666666667#0.4#0.4#>>>

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 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
–
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
–
@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
–
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
–
@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.

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## Comments

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TopNewestThere are two possibilities:

1)The capacitor charge stays the same2)The capacitor energy stays the sameIt 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 decreases2)The energy stored in the capacitor decreases3)The capacitance increases, so the time constant $\tau = R C$ increasesAnswer $D$ 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.

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

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@Steven Chase @Karan Chatrath Thanks in advance

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

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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.

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

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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?

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@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?

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

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$x = - E$

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

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Now, I am. Very interested to see how you will approach the problem.

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