Determine the charge now on each capacitor

[tiggrulz13]

Homework Statement

Two identical capacitors are connected in parallel and each acquires a charge Q₀ when connected to a source of voltage V₀. The voltage source is disconnected and then a dielectric (K=3.1) is inserted to fill the space between the plates of one of the capacitors. Assume that the capacitor without the dielectric is the first and the capacitor with the dielectric is second. Determine the charge now on each capacitor and the voltage now across each capacitor.

Homework Equations

Q = KQ₀
V = V₀/K

The Attempt at a Solution

The answers are wanted in terms of Q₀ and V₀, but I can't figure out what to do. The only solution I could come up with is Q = 3.1Q₀ and V=(1/3.1)V₀, both of which are wrong. I am also not sure how to approach the capacitor without the dielectric.

 

[AEM]

Just a quick question: Does your text use $$\epsilon = \kappa <br /> \epsilon_o$$ as the expression for the dielectric constant where $$\epsilon_o$$ is the permittivity of free space?

 

[tiggrulz13]

Yes it does. Do I need to use that?

 

[LowlyPion]

How did the charge change if the voltage source was disconnected?

 

[tiggrulz13]

I considered that and tried to just put 1Q₀ as the answer, but that was wrong.

 

[LowlyPion]

I considered that and tried to just put 1Q₀ as the answer, but that was wrong.

The charge on both will still be whatever charge they acquired when the voltage was present. The charge didn't go anywhere. It couldn't because it was disconnected.

But the capacitors are still connected in ||

 

[LowlyPion]

That means that the charge will redistribute itself over the two capacitors.

Figure the new equivalent capacitance and that total charge 2Q_o is now distributed according to the values of the new capacitance.

The capacitor with a dielectric has an increased capacity of 3.1C_o so total capacitance is 4.1C_o ...

 

[tiggrulz13]

Ok, so I found the two values for Q₁ and Q₂ (.49 and 1.51, respectively) and used that to solve for voltage. Thanks!