Electric Field of Nonuniform Line of Charge

[soccerj17]

Homework Statement

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.

Homework Equations

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

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?

 

[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.

 

[tomcue]

I would like to clarify that the equation for electric field strength in this situation is E=\int(kdq/r^2), where dq is the small charge element along the length of the rod. In this case, dq=\lambda_{0}*(x/L)^{2}dl, where dl is a small length element along the rod.

To solve for the electric field at x=-0.19m, we can use the hint given and substitute u=x+L into the integral equation. This will give us E=\int(k\lambda_{0}(u/L)^2du)/u^2 from u=L to u=0.

To find the value of this integral, we can use the fundamental theorem of calculus and evaluate the integral as E=(k\lambda_{0}/L^2)*[1/u] from u=L to u=0. Substituting in the values for L and \lambda_{0}, we get E=-(8.99*10^9)(40*10^-6)/(0.19)^2= -8.61*10^5 N/C.

Therefore, the electric field at x=-0.19m is -8.61*10^5 N/C, pointing in the negative x direction. This negative sign indicates that the electric field is indeed pointing towards the negatively charged point.