# Potential due to a rod

1. Oct 29, 2005

### thenewbosco

A rod of length L lies along the x-axis with the left end at the origin. It has non-uniform charge density given by $$\lambda = \alpha x$$. where $$\alpha$$ is a positive constant. Calculate the potential at a position A, which is a distance d to the left of the origin.

Here is what i did:

we have $$dV = \frac{k_e dq}{r}$$

then i made use of the $$\lambda = \alpha x$$

and we have $$V=\int\frac{k_e\lambda dx}{x} = \int\frac{k_e\alpha x dx}{x} = k_e\alpha (d+L-d)$$ where my limits of integration were L to d+L in the second integral

however the answer provided says that $$V=k_e\alpha (L-d ln(1 + \frac{L}{d})$$.

where did i go wrong?

thanks

Last edited: Oct 29, 2005
2. Oct 29, 2005

### Staff: Mentor

$$V=\int\frac{k_e\lambda dx}{x} = \int\frac{k_e\alpha x dr}{r}$$

then determine the relationship between r and x, i.e. r = r(x), which will also determine the relationship between dr and dx.

x is the distance along the rod, r is the separation between the incremental charge dq and the point at which the potential is being measured.

3. Oct 29, 2005

### thenewbosco

so i should put r = d + x???

4. Oct 29, 2005

### Staff: Mentor

Given: A rod of length L lies along the x-axis with the left end at the origin.

Calculate the potential at a position A, which is a distance d to the left of the origin.

Perhaps one can draw a diagram.

x [0,L]
d to left of origin
r [d,L+d] = d+x

so what is dr if 'd' is constant?

5. Oct 30, 2005

### thenewbosco

if d is constant dr is just d+x, this is correct right? and i have a diagram that came with the question but i still dont see what to put into the integral or how...

6. Oct 30, 2005

### thenewbosco

ok i got it it was really simple actually.

7. Oct 30, 2005

### Tide

Incidentally, it's not the "potential due to a rod." It's the electrical potential due to electrical charge that happens to reside on a rod.