# Non uniform charge density and electric potential

1. Jun 12, 2012

### usfelectrical

As shown in the figure, a rod of length 9.8 m lies along the x-axis, with its left end at the origin. The rod has a non-uniform linear charge density λ = αx, where α = 0.009 C/m2 and x is the position. Point A lies on the x-axis a distance 3.59 m to the left of the rod, as shown in the figure.
Q is the total charge on the rod. The Coulomb constant is 8.988 × 109 N m2/C2.

Find the potentia at A, and answer in units of volts

the two relevant equations i can find are:
v=kq/r
dv=kdq/r and lambda=qx?

i am completely stuck on this problem and just need to find a way to start it. I though that you might be able to integrate the charge density formula over the total area of the rod and treat that value as Q and then plug that value into v=kq/r where r is the 3.59, but apparently that is worng and i am back to square one with no idea as to what to do

2. Jun 12, 2012

### ehild

The potential is additive, you need to integrate the contribution dV of a small segment of the rod dx with charge dq=λdx for the whole rod:
V(A)=∫(kdq/r)=∫(kλdx/r).

ehild

#### Attached Files:

• ###### chargedrod.JPG
File size:
3.4 KB
Views:
493
Last edited: Jun 12, 2012
3. Jun 12, 2012

### usfelectrical

wait so then what would be the values for the r? cause i have been using just 3.95 since thats the distance of your point away from the rod but it doesn't work

4. Jun 12, 2012

### Staff: Mentor

Maybe this will help: r = 3.59 + x

5. Jun 12, 2012

### usfelectrical

ok, so then in the last integral you will have a ln, but would that be the same that you integrate over?

6. Jun 12, 2012

### Staff: Mentor

This equation is wrong. It should be dq = $\lambda$x dx.

Integrate this over the total length of the rod to get $\lambda$ expressed in terms of q. Then integrate dv=kdq/r to get total v at A.

7. Jun 12, 2012

### usfelectrical

i know that you have to integrate dq over the rod, what I'm saying is what do you integrate dv over?

8. Jun 12, 2012

### ehild

dV is the contribution from dq to the potential at A (with respect to infinity). The integral of dV is simply V(A) the potential at A. Without charge, V is zero.

ehild

9. Jun 13, 2012

### usfelectrical

ok, so the integral for dv would be from 0 to 3.59 because you're integrating to the potential at A?

10. Jun 13, 2012

### ehild

dV is in volts, not in meters. dV is integrated over the potential belonging to zero charge to the actual potential V(A) of the total charge of the rod. The integral on the right-hand side goes along the length of the rod, from x= to x=9.8 m.
You need to use the 3.59 m in the term "r", distance from the charge dq to A.

ehild

Last edited: Jun 13, 2012