# Capacitance Homework Word Problem

• mer584
In summary, the problem involves two capacitors, a 2.50uF capacitor charged to 857V and a 6.80uF capacitor charged to 652V, being disconnected from their batteries and then connected to each other in parallel. Using the formula Q=CV, the charge on each capacitor is found to be Q1 = .0021C and Q2= .0044 C. To find the potential difference across both capacitors, the capacitance of the two capacitors in parallel is computed and then used in the formula Q=CV. Finally, using the same formula with the original capacitors, the charge on each is determined.
mer584
The problem.
A 2.50uF capacitor is charged to 857V and a 6.80uF capacitor is charged to 652V. These capacitors are then disconnected from their batteries. Next the positive plates are connected to each other and the negative plates are connected to each other. What will be the potential difference across each and the charge on each? (note that charge is conserved)

## Homework Equations

PE= V/Q ; V=Ed; Q=CV; C= Eo (A/D); Vb-Va

Attempt:
What I started problem I used the forumla Q=CV to find the charge that is occurring in each case finding Q1 = .0021C and Q2= .0044 C. I really wasnt sure if this was even helpful or where to go from there. It appears to be a uniform field so I know I can use V=Ed but we don't have a distance or an area in this problem.

In order to get the potential difference I know you need to work with Vb-Va and possibly the potential energy. Should I use PE= V/Q then subtract the potential energies to find the work and then the poential difference?

I would compute the capacitance of the two capacitors in parallel, then use Q=CV for this capacitor to get the potential difference across both, and then use Q=CV for the 2 origional capacitors to get the charge on each

Dear student,

Thank you for providing your attempt at the problem. Let's work through it together.

First, we know that the charge on each capacitor must be conserved when they are connected in parallel. This means that the total charge on both capacitors before and after connection must be the same. We can set up an equation to represent this:

Q1 + Q2 = Q1' + Q2'

Where Q1 and Q2 are the charges on the capacitors before connection, and Q1' and Q2' are the charges on the capacitors after connection.

Next, we can use the formula Q=CV to find the charges on each capacitor before connection. Plugging in the given values, we get:

Q1 = (2.50uF)(857V) = 2142.5uC
Q2 = (6.80uF)(652V) = 4433.6uC

Substituting these values into our equation, we get:

2142.5uC + 4433.6uC = Q1' + Q2'

Next, we need to find the potential difference across each capacitor after connection. Since they are connected in parallel, they will have the same potential difference. We can use the formula V=Ed to find this.

The electric field (E) is the same for both capacitors since they are in the same uniform field. We can find this field by dividing the potential difference by the distance between the plates. However, since we don't have the distance, we can use the fact that the electric field is uniform and therefore the potential difference is proportional to the distance. This means that the potential difference will be the same for both capacitors after connection as it was before connection. So we can use the given potential differences to find the potential difference after connection:

V = 857V - 652V = 205V

Now, we can use the formula V=Ed to find the electric field:

205V = Ed

We don't have the distance, but we know the area (A) and the electric constant (Eo). We can rearrange the formula to solve for the distance:

d = V/(Eo*A)

Plugging in the values, we get:

d = (205V)/((8.85x10^-12 C^2/N*m^2)(0.0025m^2)) = 9.27x10^-6 m

## What is capacitance?

Capacitance is the ability of a system to store an electric charge.

## What is a capacitor?

A capacitor is an electronic component designed to store electric charge. It is made of two conductive plates separated by an insulating material.

## How do you calculate capacitance?

Capacitance is calculated by dividing the amount of charge stored on a capacitor by the potential difference (voltage) between the plates.

## What units are used to measure capacitance?

Capacitance is measured in farads (F), named after the English physicist Michael Faraday.

## How does capacitance affect an electric circuit?

Capacitance affects an electric circuit by storing and releasing energy, which can affect the flow of current and the overall behavior of the circuit.

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