Question:
A thin, homogeneous rod of length $l$ has the charge density $q$ (charge per length unit). The force between two point charges $Q_1$ and $Q_2$ at distance $r$ is equal to:
$$k\frac{Q_1Q_2}{r^2}$$
A) Calculate the force by which the rod attracts a charge Q of opposite sign, located at a distance $a$ from the end of the rod.
B) How large is the attractive force if we instead place the charge $Q$ in a normal plane through the rods center point at distance $\frac{l}{2}$ from the rod?
Attempted solution:
A)
First, we draw an image of the situation:
We put down a coordinate system with 0 at the left end of the rod. The rod is of length $l$ and then there is another distance $a$ to the point charge.
Let us divide the rod into very small segments that we call $dx$.
Then we create an expression for $dF$ where $q$ is the charge of the rod:
$$dF = \frac{Qq}{(L+a-x)^2} dx$$
The force then becomes:
$$F = \int_a^l \frac{kQq}{(L+a-x)^2} dx = \Big[ \frac{kQq}{L+a-x} \Big]_0^l = \frac{kQql}{a(a+l)}$$
B)
Here I am not sure I understand the setup. Is the point charge located "above" the center of the rod? Similar to the location of the letters "dx" in the image above? Over what would be integrated? It does not seem to be a one-dimensional problem?
The expected answer is:
$$\frac{k2\sqrt{2}Qq}{l}$$
The appearance of the square root of 2 leads me into thinking that there are some trigonometric things going on.
Any suggestions for how to finish this question off?
Best Answer
Due to symmetry, the net attractive force is along the vertical direction. Then,
$$dF =\frac{kQq\cos\theta}{r^2}dx = \frac{kQq\frac L2}{(\frac{L^2}4+x)^{3/2}} dx$$
and the total force below can be integrated with the substitution $t= \frac{2x}L$,
$$F = \int_{-\frac L2}^{\frac L2} \frac{kQq\frac L2}{(\frac{L^2}4+x)^{3/2}} dx =\frac{2kQq}L \int_{-1}^1 \frac{dt}{(1+t^2)^{3/2}}=\frac{2\sqrt{2}kQq}L$$