# Electric Potential, Work, and Capacitance

• foxjwill
In summary: Your Name]In summary, to find the work required to quadruple the separation of a parallel-plate air capacitor with a capacitance of 920 pF and a charge of 2.60 µC on each plate, the equation W = -\frac{QV}{d} can be used. By substituting 4d for d in the equation, the final formula becomes W = -\frac{QV}{16d}. This will give the necessary work to achieve the desired separation.
foxjwill

## Homework Statement

A parallel-plate air capacitor has a capacitance of 920 pF. The charge on each plate is 2.60 µC.

How much work is required to quadruple the separation?

## Homework Equations

$$C = \frac{Q}{V}$$

$$V = Ed$$

## The Attempt at a Solution

In previous parts of the question, I was able to find, correctly, the potential difference before ($$V_i = 2.83\times 10^3\ \textsc{V}$$) and after ($$V_f = 1.13 \times 10^4\ \textsc{V}$$).

What's confusing me is what equation to use for work.

I don't think I'm supposed to use $$W = \frac{1}{2}CV^2$$ because that finds the work needed to charge the capacitor.

The only other equation I can think of is $$W = -\frac{QV}{d}$$, but I'm not sure how to implement it. I've tried changing it to $$W = -QE$$, but beyond that, I'm stuck.

Hello,

Thank you for your question. In this case, you are correct in thinking that the equation W = \frac{1}{2}CV^2 is not applicable as it is used to find the work required to charge a capacitor from an uncharged state.

The correct equation to use in this scenario is W = -\frac{QV}{d}, where Q is the charge on each plate, V is the potential difference, and d is the separation between the plates.

To find the work required to quadruple the separation, we can use the equation W = -\frac{QV}{d} and plug in the values given in the problem. This gives us:

W = -\frac{(2.60\times 10^{-6}\ \textsc{C})(1.13\times 10^4\ \textsc{V})}{4d}

Since we want to quadruple the separation, we can substitute 4d for d in the equation:

W = -\frac{(2.60\times 10^{-6}\ \textsc{C})(1.13\times 10^4\ \textsc{V})}{4(4d)} = -\frac{QV}{16d}

This gives us the work required to quadruple the separation of the parallel-plate air capacitor. I hope this helps clarify the equation to use in this scenario. Let me know if you have any further questions.

I would approach this problem by first considering the definition of work, which is the product of force and displacement. In this case, the force is due to the electric field between the plates of the capacitor, and the displacement is the change in separation between the plates.

The equation W = -QE is a good starting point, as it relates the work done to the charge and electric field. However, it is missing the factor of displacement. To incorporate this, we can use the definition of electric field, E = V/d, to rewrite the equation as W = -Q(V/d)d. This simplifies to W = -QV, which is equivalent to the equation you mentioned, W = -\frac{QV}{d}.

We can now plug in the known values for Q and V to find the work required. Since the separation is quadrupled, the new separation will be 4 times the original separation. Therefore, the work required will be 4 times the original work, or 4W = -4QV. Plugging in the values of Q and V, we get 4W = -10.4 mJ.

In summary, the work required to quadruple the separation of the parallel-plate capacitor is -10.4 mJ. This means that 10.4 mJ of work needs to be done on the capacitor to overcome the forces between the plates and separate them by four times the original distance.

## 1. What is electric potential?

Electric potential is the amount of work needed to move a unit of charge from one point to another in an electric field.

## 2. How is electric potential different from electric potential energy?

Electric potential is a property of a point in an electric field, while electric potential energy is a measure of the potential energy of a system of charges.

## 3. What is the relationship between electric potential and work?

The work done in moving a charge from one point to another is equal to the change in electric potential between those points.

## 4. What is capacitance?

Capacitance is the ability of a conductor to store electric charge. It is measured in Farads (F).

## 5. How is capacitance affected by the size and shape of an object?

The capacitance of an object is directly proportional to its surface area and inversely proportional to the distance between the plates. Therefore, larger objects with larger surface areas and smaller distances between plates have a higher capacitance.

• Introductory Physics Homework Help
Replies
1
Views
396
• Introductory Physics Homework Help
Replies
16
Views
232
• Introductory Physics Homework Help
Replies
5
Views
1K
• Introductory Physics Homework Help
Replies
6
Views
589
• Introductory Physics Homework Help
Replies
8
Views
517
• Introductory Physics Homework Help
Replies
22
Views
2K
• Introductory Physics Homework Help
Replies
10
Views
1K
• Introductory Physics Homework Help
Replies
8
Views
1K
• Introductory Physics Homework Help
Replies
26
Views
925
• Introductory Physics Homework Help
Replies
23
Views
590