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Point Charges on a Polygon with another Charge in the Middle

CDL
20
1
1. Homework Statement
Suppose we have a regular n-gon with identical charges at each vertex. What force would a charge ##Q## at the centre feel? What would the force on the charge ##Q## be if one of the charges at the vertices were removed?



2. Homework Equations

Principle of Superposition, the net force on the charge at the centre is the sum of the forces due to the charges at each vertex.

Coulomb's Law, force on charge Q by charge q, ## \textbf{F} = \frac{qQ}{4 \pi \epsilon_0 r^2} \hat{\textbf{r}}## where ##\textbf{r}## points from q to Q, ##r = ||\textbf{r}||##.



3. The Attempt at a Solution

Construct the n-gon by placing a charge ##q_0## with charge ##q## on the positive x-axis, distance ##r## from the origin, then placing the rest of the charges ##q_1,\dots,q_{n-1}## such that they are equally spaced along the circumference of a circle of radius ##r##.

The position in polar coordinates of the ##m^{th}## charge is ##(r \text{cos}(\frac{2 \pi m}{n}), r \text{sin}(\frac{2 \pi m}{n}) )##

The total force on the charge Q is $$\textbf{F} = \sum_{m=0}^{n-1} \textbf{F}_m$$ where $$\textbf{F}_m = \frac{qQ}{4 \pi \epsilon_0 r^2} \cdot \left(-\left(\text{cos}\left(\frac{2 \pi m}{n}\right), \text{sin}\left(\frac{2 \pi m}{n}\right)\right) \right)$$ We have the ##-## sign since we require that the vector points from the vertex to the origin, by Coulomb's law.

Hence, $$\textbf{F} = \sum_{m=0}^{n-1} -\frac{qQ}{4 \pi \epsilon_0 r^2} \cdot \left( \text{cos}\left(\frac{2 \pi m}{n}\right), \text{sin}\left(\frac{2 \pi m}{n}\right) \right)$$ Therefore $$\textbf{F} = -\frac{qQ}{4 \pi \epsilon_0 r^2} \cdot \left(\sum_{m=0}^{n-1} \text{cos}\left(\frac{2 \pi m}{n}\right),\sum_{m=0}^{n-1} \text{sin}\left(\frac{2 \pi m}{n}\right) \right) = \textbf{0}$$ As the sums inside the vector components are zero. Hence, the charge ##Q## at the origin feels no force.

Now, we remove the ##k^{th}## charge. The effect that this has on the force, is that ##\textbf{F}_k## is no longer present in the sum. Hence, we have $$\textbf{F} = \left(\sum_{m=0}^{n-1} \textbf{F}_m\right) - \textbf{F}_k = -\textbf{F}_k = -\frac{qQ}{4 \pi \epsilon_0 r^2} \cdot \left(-\left(\text{cos}\left(\frac{2 \pi k}{n}\right), \text{sin}\left(\frac{2 \pi k}{n}\right)\right) \right)$$ Which results in $$\textbf{F} = \frac{qQ}{4 \pi \epsilon_0 r^2} \cdot \left(\text{cos}\left(\frac{2 \pi k}{n}\right), \text{sin}\left(\frac{2 \pi k}{n}\right)\right) $$ So, the net result is as if there were a charge identical to the one removed opposite the one that was removed. Is this correct?
 

Answers and Replies

Nathanael
Homework Helper
1,650
238
I didn’t check your math but I think your answers are correct.

I think it could’ve been much simpler by symmetry arguments, maybe something like this:
“The n-gon is unchanged by rotations of 360/n degrees. The net force should also be unchanged by this rotation (because the system is identical) and therefore must be the zero vector.”
As for part 2, the principle of superposition says you can use the result from part 1 and then just add an opposite charge to one of the vertices (to make that vertex chargeless).
 
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