# J"Calculating Work for Charging a Parallel Plate Capacitor

• xswtxoj
In summary, a parallel plate capacitor with an electric potential of 334V is charged by moving 3.78e16 electrons from one plate to the other. The work required to charge the capacitor can be calculated using the formula E=0.5*Q*V, where Q is the charge in Coulombs and V is the potential difference. However, the rearrangement of the formula in the given solution is incorrect and the charge should be converted from electron charges to Coulombs before being used in the formula.
xswtxoj

## Homework Statement

A parallel plate capacitor is changed to an electric potential of 334V by moving 3.78e16 electrons from one plate to the other. How much work is charging the capacitor?

## Homework Equations

Delta V= Delta PE/ q
c=q/v

## The Attempt at a Solution

Delta V= Delta PE/ q
rearrange to q * PE = 3.78E16 *334= 1.26E19

What fraction of a coulomb is the 3.78*1016 electrons of charge?

I don't like the look of your "Delta V= Delta PE/ q" in this case. Due to the fact that the V changes as charge is added to the capacitor, you get a slightly more complicated formula for the energy of the charged capacitor: E = 0.5*Q*V. Roughly speaking, you use the average V so you get a factor of 0.5. Check it out at http://en.wikipedia.org/wiki/Capacitor

Delta V= Delta PE/ q
rearrange to q * PE = 3.78E16 *334= 1.26E19
Here you have rearranged incorrectly. Also, you have entered the charge in non-standard units. The charge is given as a certain number of electron charges, and this number must be multiplied by the charge in Coulombs on one electron.

i think i should be the same where 1 j/c is = to 1 Ev, so ur saying the formula can't be done? (E=.5*q*v)

## 1. What is the equation for calculating work for charging a parallel plate capacitor?

The equation for calculating work for charging a parallel plate capacitor is W = 1/2 * C * (Vf^2 - Vi^2), where W is the work done, C is the capacitance, and Vf and Vi are the final and initial voltages, respectively.

## 2. How do you determine the capacitance of a parallel plate capacitor?

The capacitance of a parallel plate capacitor can be determined by the equation C = ε0 * A / d, where ε0 is the permittivity of free space, A is the area of the plates, and d is the distance between the plates.

## 3. Can the work done in charging a parallel plate capacitor ever be negative?

No, the work done in charging a parallel plate capacitor can never be negative as it requires energy to separate the charges and store them on the plates. However, the change in energy of the capacitor may be negative if the final voltage is lower than the initial voltage.

## 4. How does the distance between the plates affect the work done in charging a parallel plate capacitor?

The distance between the plates has an inverse relationship with the work done in charging a parallel plate capacitor. As the distance increases, the work done decreases, and vice versa. This is because a larger distance between the plates results in a smaller capacitance, which requires less work to charge.

## 5. What is the unit of work when calculating the work for charging a parallel plate capacitor?

The unit of work when calculating the work for charging a parallel plate capacitor is joules (J), which is the unit of energy. This is because work is the energy transferred to or from the capacitor during the charging process.

• Introductory Physics Homework Help
Replies
18
Views
2K
• Introductory Physics Homework Help
Replies
6
Views
451
• Introductory Physics Homework Help
Replies
1
Views
273
• Introductory Physics Homework Help
Replies
11
Views
568
• Introductory Physics Homework Help
Replies
20
Views
1K
• Introductory Physics Homework Help
Replies
8
Views
1K
• Introductory Physics Homework Help
Replies
58
Views
3K
• Introductory Physics Homework Help
Replies
4
Views
509
• Introductory Physics Homework Help
Replies
3
Views
1K
• Introductory Physics Homework Help
Replies
20
Views
2K