Five identical point charges of positive charge $q$ are placed at five vertices of a regular hexagon of side length $a$. A point charge carrying charge $-q$ is placed at the center, as shown in the figure below.

Use the $\left(x,y\right)$ coordinate system provided in the figure.

1. Based on symmetry, determine the direction of the net force exerted on the negative point charge.

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We begin by numbering the point charges as follows for convenience.

We then draw the force exerted by each point charge on the charge located at the center and realize that because all of the five point charges have the same charge $q$ and are located the same distance $a$ from the central charge, the forces ${\overrightarrow{F}}_1$ and ${\overrightarrow{F}}_4$ cancels as well as the forces ${\overrightarrow{F}}_2$ and ${\overrightarrow{F}}_5$.

Thus, the only force that doesn’t have an opposing force to cancel with is the force ${\overrightarrow{F}}_3$ and we conclude that the net force acting on the negative point charge is equal to ${\overrightarrow{F}}_3$ and must point to the left because the negative point charge will be attracted to the left most positive charge.

2. Determine the magnitude of the net force and write the force in vector notation.

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From our previous answer, we know that the only force that doesn’t have an opposing force to cancel with is the force ${\overrightarrow{F}}_3$ and we conclude that the net force acting on the negative point charge points to the left and is given by

\boxed{{\overrightarrow{F}}_{net}={\overrightarrow{F}}_3=-\frac{kq^2}{a^2}\ \ \hat{x}}

3. Now, you additionally place a negative point charge $-q$ at every vertex of the regular hexagon. Determine the magnitude and direction of the net force acting on the point charge located at the center.

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Adding a charge $-q$ to each vertex is equivalent to have a zero net charge on vertices 1 through 5 and a sole negative charge $-q$ on vertex 6.

Thus, the vertices 1 through 5 will not exert a force on the central negative charge and only the charge $-q$ on vertex 6 will create a repulsive force on it. The net force acting on the point charge located at the center will therefore be equal to

\boxed{{\overrightarrow{F}}_{net}={\overrightarrow{F}}_6=-\frac{kq^2}{a^2}\ \ \hat{x}}

4. Compare your answers to parts 2. and 3. and explain.

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In both cases the net force acting on the point charge located at the center is the same. This comes from the fact that the first distribution, in terms of net force at the center, is equivalent to a single point charge $q$ on vertex 3 which attracts the central negative point charge. The second distribution, in terms of net force at the center, is equivalent to a single point charge $-q$ on vertex 6 which repels the central negative point charge. Both distributions therefore create the same force on the point charge located at the center.

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