Work against an electric field due to a point charge

[mhen333]

Homework Statement

I need help seeing if I did this right...

This is from Fundamentals of Physics, 8th edition, volume 2. By Jearl Walker.
Chapter 24, problem 88.

A particle of positive charge Q is fixed at point P. A second particle of mass m and negative charge -q moves at a constant speed in a circle of radius $$r_{1}$$ centered at P. Derive an expression for the work W that must be done by an external agent on the second particle to increase the radius of the circle of motion to $$r_{2}$$.

Homework Equations

$$<br /> W_{\vec{E}} = -\int_{r_{1}}^{r_{2}} \vec{E}\cdot d\vec{r}<br />$$
$$\vec{E} = \frac{1}{4\pi\epsilon_{0}} \frac{Q}{r^2} \hat{r}$$

The Attempt at a Solution

Here's what I'm thinking. The work against the electric field is just the integral above. Since the particle stays on an equipotential surface, I should only have to worry about work outward, right?

Here's what I have:

$$<br /> W_{total} = W_{\vec{E}} = \left(\frac{1}{r_{2}} - \frac{1}{r_{1}}\right) \left(\frac{Q}{4\pi\epsilon_{0}}\right)<br />$$

Please help point out if I've done something wrong!-Mike

 

[AEM]

Your idea is correct, but your first equation is wrong. Usually gravity is ignored because the gravitational force is small compared to the electric force.

 

[mhen333]

I think I figured out where I went wrong in my equation.

So I'm confused, where does the mass 'm' come in?

 

[AEM]

In brief, Work = Force times distance. So your equation for work should be

$$W_{\vec{E}} = -\int_{r_{1}}^{r_{2}} q \vec{E}\cdot d\vec{r}$$

Where the force is qE, q being the charge that's moved, and E is the field due to the charge Q.

 

[AEM]

Also, I should mention that you are correct. Work is done only when the charged is moved along a radial path.

 

[AEM]

Besides the approach that I mentioned in my previous post, you can also use the fact that Work = qV where q is the charge that's being moved, and V is the change in electric potential.

What you calculated in your work below, was the change in electric potential, not the work. All you have to do is change the "W" to "V" and multiply through by the charge q.

You were on the right track. Just forgot the definition, W = qV. So now you have two ways to look at it.

Homework Equations

$$<br /> W_{\vec{E}} = -\int_{r_{1}}^{r_{2}} \vec{E}\cdot d\vec{r}<br />$$
$$\vec{E} = \frac{1}{4\pi\epsilon_{0}} \frac{Q}{r^2} \hat{r}$$

The Attempt at a Solution

Here's what I'm thinking. The work against the electric field is just the integral above. Since the particle stays on an equipotential surface, I should only have to worry about work outward, right?

Here's what I have:

$$<br /> W_{total} = W_{\vec{E}} = \left(\frac{1}{r_{2}} - \frac{1}{r_{1}}\right) \left(\frac{Q}{4\pi\epsilon_{0}}\right)<br />$$

Please help point out if I've done something wrong!


-Mike

 

[mhen333]

Thank you so much!

I've been nearly pulling my hair out over this.