Point Charges on a Polygon with another Charge in the Middle

In summary: So the force on the center is the force on Q from this new charge (which is the same as from the charge that was there before) plus the force on Q from the charge that was added.In summary, the net force on a charge at the center of a regular n-gon with identical charges at each vertex is zero. This can be determined using Coulomb's Law and the principle of superposition. If one of the charges at the vertices is removed, the net force on the center charge will be equal to the force from the charge that was removed, plus the force from the added charge that balances out the removed charge.
  • #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]
 
Physics news on Phys.org
  • #2
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).
 
  • Like
Likes CDL

Related to Point Charges on a Polygon with another Charge in the Middle

1. How does the number of sides on the polygon affect the electric field at the center?

The number of sides on the polygon has no effect on the electric field at the center. The electric field at any point due to a collection of point charges is determined by the distance from the point charge and the magnitude of the charge, not the shape of the configuration.

2. Can the electric field at the center of the polygon ever be zero?

Yes, the electric field at the center of the polygon can be zero if the charges on the polygon are symmetrically arranged or if the total magnitude of the charges on the polygon is equal to the magnitude of the charge in the center.

3. Does the distance between the point charges on the polygon affect the electric field at the center?

Yes, the distance between the point charges on the polygon does affect the electric field at the center. As the distance between the charges decreases, the electric field at the center increases, and vice versa.

4. How does the magnitude of the charge in the center affect the electric field at the center of the polygon?

The magnitude of the charge in the center directly affects the electric field at the center of the polygon. As the magnitude of the charge increases, the electric field at the center also increases, and vice versa.

5. Can the electric field at the center of the polygon be negative?

Yes, the electric field at the center of the polygon can be negative. This occurs when the charges on the polygon are arranged in such a way that the net electric field at the center is directed opposite to the direction of the electric field due to the charge in the center.

Similar threads

  • Introductory Physics Homework Help
Replies
11
Views
850
  • Introductory Physics Homework Help
Replies
6
Views
2K
  • Introductory Physics Homework Help
Replies
5
Views
1K
  • Introductory Physics Homework Help
Replies
12
Views
411
  • Introductory Physics Homework Help
Replies
6
Views
220
  • Introductory Physics Homework Help
Replies
23
Views
533
  • Introductory Physics Homework Help
Replies
5
Views
654
  • Introductory Physics Homework Help
Replies
10
Views
299
  • Introductory Physics Homework Help
Replies
6
Views
292
  • Introductory Physics Homework Help
Replies
5
Views
3K
Back
Top