- #1
CDL
- 20
- 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? [/B]
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}||##.[/B]
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?[/B]