# Work done by moving a point-charge

• AwesomeTrains
In summary, the homework statement is to calculate the work needed to move a point-charge with the charge q from infinity to the center of the chargedistribution given by:\rho(\vec{r})=\rho_0 e^{-\alpha r}
AwesomeTrains
Hello everyone!

1. Homework Statement

Calculate the work needed to move a point-charge with the charge q from infinity to the center of the chargedistribution given by:
$\rho(\vec{r})=\rho_0 e^{-\alpha r}$

## Homework Equations

$U=qV$
$V=\frac{1}{4\pi\epsilon_0}\int_{V}\frac{\rho(\vec{r})}{r}dV$

## The Attempt at a Solution

I did the integral by using integration by parts:
$\frac{1}{4\pi\epsilon_0}\int_{V}\frac{\rho(\vec{r})}{r}dV=\frac{1}{4\pi\epsilon_0}\int_0^{\infty}\int_0^{2\pi}\int_0^{\pi}\frac{\rho_0 e^{-\alpha r}}{r}r^2=\frac{\rho_0}{\epsilon_0 \alpha^2}$
The work should then be, for moving the charge from infinity to 0, $U=\frac{\rho_0 q}{\epsilon_0 \alpha^2}$
Is this correct? Please let me know if I should elaborate some steps.

You cannot use V for volume and for voltage in the same equation.
In the equation for potential at position ##\vec r##, V is not going to be the entire volume enclosing the charge.
The potential energy wanted is the work needed to move the charge from infinity to the origin.
How would you figure the potential energy to put the same charge a distance R from the origin?

I think I would do a path integral over $\vec{E}$.
$-\int_0^Rq\vec{E}\cdot d\vec{S}$.
To do that I'm trying to calculate the field, I got stuck at that though.
$\vec{E}(\vec{r})=\frac{1}{4\pi\epsilon_0}\int_{volume}d\vec{r}'\rho_0 e^{-\alpha r'}\frac{\vec{r}-\vec{r}'}{|\vec{r}-\vec{r}'|^3}$
How do I integrate the fraction in the integrand? I have to integrate over the whole space in spherical coordinates?

... so first off I notice that you would not use the potential equation like you did before. Why not? What changed in your reasoning?

But yes - by definition, the change in potential energy between point P and point Q is given by the path integral ##\Delta U = \int_P^Q \vec F \cdot d\vec s##
For a conservative force, this equation becomes ##\Delta U = U(Q)-U(P)## - you've seen these equations before right?
... also by definition, the potential energy at a point is this integral, but P is at infinity.

Notes: for a spherically symmetric charge distribution:
The charge below radius r is ##Q(r)=4\pi\int_0^r \rho(r')r'^2\;dr'## (check);
The potential at radius r is ##\phi(r)=Q(r)/r## where ##U=q\phi## ... equipotential surfaces are spherical shells.

I thought if the potential equation wasn't right in this case, I'd calculate the work in the same way as for a gravitational field.

Simon Bridge said:
$dW=q\vec{E}(\vec{r})dr$, could I then integrate r from 0 to $\infty$ and use the fact that $\phi(r)=-\nabla \vec{E}(r)$ in spherical coordinates?

Simon Bridge said:
For a conservative force, this equation becomes ΔU=U(Q)−U(P) ΔU = U(Q)-U(P) - you've seen these equations before right?
This field is conservative because it is time independent? I think we derived these equations in classical mechanics for the gravitational field.

I tried evaluating the charge below r, which gave the potential at r: $\phi(r)=\frac{4\pi\rho_0(e^{-\alpha r}(-\alpha^2 r^2 - 2\alpha r -2 ) +2)}{\alpha^3 r}$
If I then try to get the potential difference between 0 and $\infty$ I get 0: $lim_{P->\infty}U(P)=0$ and $lim_{Q->0}U(Q)=0$ then $\Delta U = U(Q)-U(P)=U(0)-U(\infty)=0$
I think I'm missing some important point.

So basically, in response to questions, you have come up with at least three different approaches, and you are as confused as ever.

One of your sticking points is not understanding the relations you use - the other may be a confusion about how you apparently have to integrate over the radius twice (once to get a Q(r) and again to get U).

I think you've shown you are OK with the actual maths part - it's just figuring what it all means. So here's the start of a walkthrough:

The potential energy at a point is the work needed to get the charge there from infinity ...
The electric field is conservative##^1##, so the path does not matter... so pick a path that makes the maths simple:
Since the force is always radially outwards, come in directly along the radius.
##\vec F(\vec r)\cdot d\vec r = -F(r)\; dr## (force points opposite to the path taken##^2##.)
So $$U = -\int_\infty^0 f(r)\;dr = \int_0^\infty F(r)\;dr$$ ... follow me so far##^3##?

Since ##d\vec F=q\; d\vec E## ... $$F(r)=\frac{q}{4\pi\epsilon_0}\frac{Q(r)}{r^2}$$... where Q(r) is the total charge inside a ball of radius r.

That would be: $$Q(r)=4\pi\int_0^r u^2\rho(u)\; du$$ ... "u" plays the role of a dummy variable radius coordinate##^4##.

--------------------------------------
[1] a conservative field is one where the total work done moving around any closed path is zero.
[2] I may have this wrong ... but it's easy to troubleshoot.
[3] dU = F(r)dr ... the work done to get q from r+dr to r is the same as that done by the force getting q from r to r+dr.
[4] I'd normally use r' in this role but I used a u-substitution u=r' because I want to make the distinction clear - in the equation for Q, "r" is a specific radius, and u is a variable on the radial axis. In the equation for U, you are summing over a range of the specific values of radius the charge passes through to get to the origin.
The ##4\pi## out the front is because the charge between r and r+dr is ##dQ=4\pi r^2\rho(r)\; dr##

AwesomeTrains
My calculation looks like this:

First I calculated the total charge below the pointcharge q at R.
$Q(R)=4\pi \int_0^R u^2\rho(u)du=4\pi\rho_0\int_0^Ru^2e^{-\alpha u}du$
I did the integral in maple and got this:
$Q(R)=\frac{4\pi\rho_0}{\alpha^3}[e^{-\alpha R}(-R^2\alpha-2R\alpha-2)+2]$

The force from the charge below acting on the pointcharge q is then given by:
$F(r)=\frac{q}{4\pi\epsilon_0}\frac{Q(r)}{r^2}=\frac{q\rho_0}{\alpha^3 \epsilon_0 }\frac{1}{r^2}[e^{-\alpha r}(-r^2\alpha-2r\alpha-2)+2]$

Moving it dr requires the work: $dU=F(r)dr$
Integrating over the path from $\infty$ to 0 then gives the total work required:
$U=\frac{\rho_0 q}{\alpha^3 \epsilon_0}\int_0^\infty (-e^{-\alpha r} \alpha^2 - 2\alpha e^{-\alpha r} r^{-1} - 2e^{-\alpha r}r^{-2}+2r^{-2})=\frac{\rho_0 q}{\alpha^3 \epsilon_0} \alpha=\frac{\rho_0 q}{\alpha^2 \epsilon_0}$ (I did this integral in maple aswell)

Why am I still getting the result from post #1? Is my reasoning equivalent to yours?

Simon Bridge said:
In the equation for potential at position ##\vec r##, V is not going to be the entire volume enclosing the charge
Why not? This is potential, not field.

My tutor showed us the correct result in case anyone wanted to know: $W=\frac{4\pi \rho_o q}{\epsilon_0 \alpha^2}$ (I forgot the $4 \pi$ from the charge $Q(r)$ below q)
Thanks for making the physics behind the calculation more clear.

## 1. What is work done by moving a point-charge?

The work done by moving a point-charge is the amount of energy transferred to or from the point-charge as it moves in an electric field. This work is equal to the product of the magnitude of the charge, the magnitude of the electric field, and the displacement of the charge.

## 2. How is work done by moving a point-charge calculated?

The work done by moving a point-charge is calculated using the formula W = qEd, where W is work (in joules), q is the magnitude of the point-charge (in coulombs), E is the magnitude of the electric field (in newtons per coulomb), and d is the displacement of the charge (in meters).

## 3. What is the relationship between work done and potential difference in a point-charge system?

In a point-charge system, the work done by moving a point-charge is equal to the change in potential energy of the charge. This means that the work done is directly proportional to the potential difference between the initial and final positions of the charge.

## 4. How does the direction of the electric field affect the work done by moving a point-charge?

The direction of the electric field affects the work done by moving a point-charge in that the work will be positive if the electric field and the displacement of the charge are in the same direction, and negative if they are in opposite directions. This is because work is a scalar quantity and is dependent on the magnitude and direction of the electric field and displacement.

## 5. Can the work done by moving a point-charge be negative?

Yes, the work done by moving a point-charge can be negative if the electric field and the displacement of the charge are in opposite directions. This means that energy is being transferred from the point-charge to the electric field, resulting in a decrease in the potential energy of the charge.

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