"A uniform line of charge with length 20.0 cm is along the x-axis, with its midpoint at x=0. Its charge per length is +4.80 nC/m. A small sphere with charge -2.00 micro coulombs is located at x=0,y=5.00 cm. What are the magnitude and direction of the force that the charged sphere exerts on the line of charge?"

So before I get into what confused me about this problem let me show you how I solved it. I calculated the electric field from the line of charge at the center of the sphere, by deriving the following formula $$\vec{E}_{Q on Sphere}=\frac{1}{4\pi\epsilon_0}\frac{Q}{y\sqrt{\frac{L^2}{4}+y^2}}\hat{j}$$ where Q is the total charge along the line of charge, L is the length of the line of charge and y is the distance of the sphere from the line of charge, along the y-axis. I then calculated the electric field by plugging the appropriate values of L and y given in the problem and I calculated Q by multiplying the charge density by the length L. Then to get the force the line of charge exerts on the sphere I used the formula, $$\vec{F}_{Q on Sphere}=q\vec{E}_{Q on Sphere}$$ Lastly I applied newton's third law, which tells me that $$\vec{F}_{Sphere on Q}=-\vec{F}_{Q on Sphere}$$ and then I had my answer. I was wondering though is it possible to calculate the electric field of the sphere on the line of charge, $\vec{E}_{Sphere on Q}$, directly. This calculation seems very different to me and the fact I couldn't calculate this was why the problem originally stumped me. The reason why is because it doesn't seem like you can set up an integral in the same way you do to derive $\vec{E}_{Q on Sphere}$ above. To show you what I mean let me show you how I set up the integral to find $\vec{E}_{Q on Sphere}$. When I derived my integral to get $\vec{E}_{Q on Sphere}$ above I expressed the electric field from an infinitesimally small unit of charge as $$dE_{Q on Sphere}=\frac{1}{4\pi\epsilon_0}\frac{dQ}{x^2+y^2}$$ From here it is easy to see how this can be turned into an integral because $dQ=\lambda dx$, where $\lambda$ is the charge density. So to get the electric field, we integrate with respect to x. We can also ignore the x-component of the electric field by symmetry so, we have $$E_{Q on Sphere}=\int dE_{(Q on Sphere) y}=\int dE_{Q on Sphere}sin \theta$$ With some basic trigonometry we see that $$sin \theta=\frac{y}{\sqrt{x^2+y^2}}$$ Putting all this together we get the following integral, $$E_{Q on Sphere}=\frac{\lambda}{4\pi \epsilon_0}\int_{-L/2}^{L/2}\frac{y}{(x^2+y^2)^{3/2}}dx$$ Now I'll try to derive the electric field of the sphere on the line of charge. I'll start by writing out the equation for electric field at an arbitrary point on the line of charge. $$E_{Sphere on Q}=\frac{1}{4\pi \epsilon_0}\frac{q}{x^2+y^2}$$ Now this is where I got stuck. The problem I have with this expression is I see no way to integrate because there is nothing you can use to relate to a differential such as dx. Am I just taking the wrong approach here, or is calculating this electric field directly not possible?

## Best Answer

You don't calculate the electric field "on the line of charge". Or on anything. The field exists at the location of the line of charge. At any point on line of charge you have an electric field produced by the sphere. This is E and not dE. What you sum are the forces on each piece of the line of charge. You have to calculate dF=Edq and sum these contributions. But pay attention to direction of each dF. They are vectors. Calculate component along the line and perpendicular to the line separately.