Field Intensity of Flat Capacitor

[fonz]

Homework Statement

Two identical flat capacitors with no dielectric material inside are connected in series to a voltage source. The electric field intensity inside the capacitors is E. If the space inside one of the capacitors is filled with a dielectric material of ε=9 how strongly did the field intensity change if:

a) the circuit was connected to the voltage source when the capacitor was filled with a dielectric

b) the circuit had been disconnected from the voltage source before the capacitor was filled with the dielectric

Homework Equations

Electric field intensity becomes reduced by:

E = E_external / ε

C = ε₀εS/d

The Attempt at a Solution

The capacitance increases by a factor of 9

Therefore the charge on the capacitor must increase by a factor of 9 therefore the charge on the other series capacitor increased by a factor of 9

Therefore the electric field inside the filled capacitor must increase by a factor of 9

Even if the above is correct what difference does having the source connected or disconnected do? I'm assuming it must be related to the fact that the two are in series therefore the charge on each must be the same.

 

[gneill]

Homework Statement

Two identical flat capacitors with no dielectric material inside are connected in series to a voltage source. The electric field intensity inside the capacitors is E. If the space inside one of the capacitors is filled with a dielectric material of ε=9 how strongly did the field intensity change if:

a) the circuit was connected to the voltage source when the capacitor was filled with a dielectric

b) the circuit had been disconnected from the voltage source before the capacitor was filled with the dielectric

Homework Equations

Electric field intensity becomes reduced by:

E = E_external / ε

C = ε₀εS/d

The Attempt at a Solution

The capacitance increases by a factor of 9

Therefore the charge on the capacitor must increase by a factor of 9 therefore the charge on the other series capacitor increased by a factor of 9

Therefore the electric field inside the filled capacitor must increase by a factor of 9

Even if the above is correct what difference does having the source connected or disconnected do? I'm assuming it must be related to the fact that the two are in series therefore the charge on each must be the same.

If the voltage source remains connected then there is a complete circuit and current can flow when conditions change (like having a dielectric inserted in one of the capacitors). That means that the charges on the capacitors can change. The total potential across the capacitors must remain the same (equal to the voltage source).

If the voltage source is disconnected before the dielectric is inserted, then there is no complete circuit and current will not flow. That means the individual charges must remain the same on the capacitors. On the other hand, there is no longer a restriction on the total potential difference across the series capacitors.

You'll have to work within those constraints in order to determine how the charges and or voltages will change in each case.