# Electric Field of Nonuniform Line of Charge

1. Sep 2, 2009

### soccerj17

1. The problem statement, all variables and given/known data
A thin rod extends along the x-axis from x=0 to x=L and carries a line charge density $$\lambda$$ = $$\lambda_{0}$$*(x/L)$$^{2}$$, where $$\lambda_{0}$$ is a constant.

If L=0.19 m and $$\lambda_{0}$$ = 40$$\mu$$C/m, find the electric field strength at the point x=-0.19 m. (Hint: the resulting integral is easily computed with the change of variable u=x+L.

2. Relevant equations
E=$$\int(kdq/r^2)$$
dq=$$\lambda_{0}$$*(x/L)$$^{2}$$dl

3. The attempt at a solution
Since the line is positively charged, the electric field should point in the negative x direction at the point x=-0.19m and I need to use the integral equation for E. The integral I set up is E=$$\int(k\lambda_{0}(x/L)^2dx)/(x+L)^2$$ from x=0 to x=L. I'm not sure this integral is correct because I don't know how to solve it and the hint says that substituting u=x+L should make it easy to solve. The main things I'm unsure of in this equation is what variables to use. For instance, in the linear charge density, it seems like the x is the distance along the line and the L is the total length of the rod. So I would need to integrate with respect to x to get the full charge of the rod. But then I'm not sure what the "r^2" distance is in terms of the variables. It has to be x along the line plus however far away the point is away from the line, right? So what should the denominator be in the integral and how do I solve the integral?
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Sep 3, 2009

### soccerj17

I tried plugging the numbers I knew into the integral and got
E=$$\int(\frac{k\lambda_{0}x^{2}dx}{0.0361(x+0.19)^{2}}$$ and taking the constants out of the equation I got E=$$\frac{k\lambda_{0}}{0.0361}$$$$\int\frac{x^{2}dx}{(x+0.19)^{2}}$$ with the limits of the integral being x=0 to x=L (or 0.19)
Does this integral seem right? Because I still don't know how to solve it.