# Determine the charge now on each capacitor

• tiggrulz13
In summary, when the voltage source is disconnected, the capacitors each acquire a charge of 3.1Q0. Then, using equivalent capacitance and the values of Q1 and Q2, the voltage across each capacitor is found to be 1.51V0 and 1.49V0, respectively.
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.

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.

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

Yes it does. Do I need to use that?

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

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

tiggrulz13 said:
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 ||

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

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

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

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

## What is the purpose of determining the charge on each capacitor?

The purpose of determining the charge on each capacitor is to understand how much electrical energy is stored in each capacitor. This information is important for designing and troubleshooting electrical circuits.

## How do you determine the charge on a capacitor?

To determine the charge on a capacitor, you need to know the capacitance of the capacitor and the voltage across it. The charge is then calculated by multiplying the capacitance by the voltage.

## What is the unit of measurement for charge on a capacitor?

The unit of measurement for charge on a capacitor is Coulombs (C). This unit represents the amount of electrical charge that has been stored in the capacitor.

## Can the charge on a capacitor be negative?

Yes, the charge on a capacitor can be negative. This occurs when the capacitor is connected in reverse polarity, causing the charge to flow in the opposite direction. However, the magnitude of the charge will still be positive.

## How does the charge on a capacitor affect its performance?

The charge on a capacitor directly affects its ability to store and release electrical energy. A higher charge can result in a higher voltage output, while a lower charge may lead to a decrease in performance.

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