Calculating Work and Potential for Moving Charges in an Electric Field

[indigojoker]

Suppose there was a charge +Q located at x=-a and a +Q charge at x=a. If i wanted to bring in a charge of -Q form infinity to x=0, what is the work that I have to do?

just wondering if i have the right expression:
$$V=-\int _{\inf} ^a \frac{1}{4 \pi \epsilon_o } \frac{Q}{r^2}dr + -\int_{\inf}^{-a} \frac{1}{4 \pi \epsilon_o} \frac{Q}{r^2}dr$$

if i wanted to move the charge at the origin to x=+2a, the work i would have to do:

$$V=-\int _{a} ^{3a} \frac{1}{4 \pi \epsilon_o } \frac{Q}{r^2}dr + -\int_{-a}^{a} \frac{1}{4 \pi \epsilon_o} \frac{Q}{r^2}dr$$

I was also wondering where the charge of the charge I'm moving comes into play, it doesn't seem like it matters for these equations?

also, why is calculating the potential the same as finding the work?

 

[learningphysics]

Yes, those look right to me. But that gives the potential difference between infinity and that point... to get the work... you need to multiply the result by -Q.

 

[learningphysics]

But have you studied potential and energy due to point charges?

You can write the result immediately without integrals...

 

[indigojoker]

how do you do that?

 

[learningphysics]

how do you do that?

The energy of the charge at infinity is 0...

The energy of the charge at x = 0, is just:

$$\frac{kQ(-Q)}{a} + \frac{kQ(-Q)}{a} = \frac{-2kQ^2}{a}$$

I'm using the formula... electric potential energy between two point charges is:

$$\frac{kq_1q_2}{r}$$

So the change in energy is negative. So the charge loses energy... ie: you do negative work on the object...

 

[indigojoker]

something weird happened and I'm not sure if it's supposed to:
$$V=-\int _{\inf} ^a \frac{1}{4 \pi \epsilon_o } \frac{Q}{r^2}dr + -\int_{\inf}^{-a} \frac{1}{4 \pi \epsilon_o} \frac{Q}{r^2}dr =0$$

I get:

$$V=-\frac{kQ}{a} + \frac{kQ}{a} =0$$

so it takes no work to move the charge there?

 

[learningphysics]

something weird happened and I'm not sure if it's supposed to:
$$V=-\int _{\inf} ^a \frac{1}{4 \pi \epsilon_o } \frac{Q}{r^2}dr + -\int_{\inf}^{-a} \frac{1}{4 \pi \epsilon_o} \frac{Q}{r^2}dr =0$$

I get:

$$V=-\frac{kQ}{a} + \frac{kQ}{a} =0$$

so it takes no work to move the charge there?

No... you have to be careful about the integrals... if you are integrating along the x-axis...

let's examine the charge at x = a, and the voltage due to that charge. (here r = x - a, where x is the coordinate of the charge being brought in... a is the coordinate of the stationary charge)

$$-\int_{\inf}^{-a} \vec{E}\cdot\vec{dr}$$

this equals:

$$-(\int_{\inf}^{a} \vec{E}\cdot\vec{dr} + \int_{a}^{0} \vec{E}\cdot\vec{dr}+\int_{0}^{-a} \vec{E}\cdot\vec{dr})$$

$$-(\int_{\inf}^{a} \frac{1}{4 \pi \epsilon_o} \frac{Q}{r^2}dr +\int_{a}^{0} \frac{1}{4 \pi \epsilon_o} \frac{Q}{r^2}dr +\int_{0}^{-a} \frac{-1}{4 \pi \epsilon_o} \frac{Q}{r^2}dr)$$

see, the last section of that integral... the field is in the opposite direction, so a minus needs to be put in... also the a to 0... and 0 to -a parts cancel each other out... so we end up with the same integral as with the other stationary charge...

I think maybe this integral is problematic because there's a discontinuity when r = 0... but it works out anyway.

 

[indigojoker]

The energy of the charge at infinity is 0...

The energy of the charge at x = 0, is just:

$$\frac{kQ(-Q)}{a} + \frac{kQ(-Q)}{a} = \frac{-2kQ^2}{a}$$

I'm using the formula... electric potential energy between two point charges is:

$$\frac{kq_1q_2}{r}$$

So the change in energy is negative. So the charge loses energy... ie: you do negative work on the object...

so the potential at x=-a is the same as the potential at x=a?

 

[indigojoker]

No... you have to be careful about the integrals... if you are integrating along the x-axis...

let's examine the charge at x = a, and the voltage due to that charge. (here r = x - a, where x is the coordinate of the charge being brought in... a is the coordinate of the stationary charge)

$$-\int_{\inf}^{-a} \vec{E}\cdot\vec{dr}$$

this equals:

$$-(\int_{\inf}^{a} \vec{E}\cdot\vec{dr} + \int_{a}^{0} \vec{E}\cdot\vec{dr}+\int_{0}^{-a} \vec{E}\cdot\vec{dr})$$

$$-(\int_{\inf}^{a} \frac{1}{4 \pi \epsilon_o} \frac{Q}{r^2}dr +\int_{a}^{0} \frac{1}{4 \pi \epsilon_o} \frac{Q}{r^2}dr +\int_{0}^{-a} \frac{-1}{4 \pi \epsilon_o} \frac{Q}{r^2}dr)$$

see, the last section of that integral... the field is in the opposite direction, so a minus needs to be put in... also the a to 0... and 0 to -a parts cancel each other out... so we end up with the same integral as with the other stationary charge...

I think maybe this integral is problematic because there's a discontinuity when r = 0... but it works out anyway.

$$-(\int_{\inf}^{a} \frac{1}{4 \pi \epsilon_o} \frac{Q}{r^2}dr -\int_{a}^{-a} \frac{1}{4 \pi \epsilon_o} \frac{Q}{r^2}dr)$$
$$=-(kQ (-1/a)-kq(2/a))= kQ 3/a$$

maybe I'm just doing the integrals wrong

 

[learningphysics]

$$-(\int_{\inf}^{a} \frac{1}{4 \pi \epsilon_o} \frac{Q}{r^2}dr -\int_{a}^{-a} \frac{1}{4 \pi \epsilon_o} \frac{Q}{r^2}dr)$$
$$=-(kQ (-1/a)-kq(2/a))= kQ 3/a$$

maybe I'm just doing the integrals wrong

The integral from a to 0 has kQ/r^2 inside the integral, but the integral from 0 to -a should have -kQ/r^2 (with the minus sign) inside the integral... the direction of the field has changed from rightward to leftward.

So you can't put it all under the same integral...

 

[indigojoker]

why couldn't i do this:

$$-(\int_{\inf}^{0} \frac{1}{4 \pi \epsilon_o} \frac{Q}{r^2}dr +\int_{0}^{-a} \frac{-1}{4 \pi \epsilon_o} \frac{Q}{r^2}dr)$$

since the field from infinity to zero is the same?