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**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?

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}||##.

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?

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?