# Charges in a polygon distribution

• mathnerd15
In summary, when 13 equal charges are placed at the corners of a polygon and a test charge is placed at the center, the net force on the test charge can be calculated using the equation (1/4piEo)(qQ/r^2). As the number of sides in the polygon approaches infinity, the charges will cancel out and the net force will be zero. Additionally, for a circle with n=3, the y component of the force will be zero due to the rotation of one point directly in the -y direction. It can also be generalized that for a circle with any number of charges, the horizontal net force at each node will be zero.
mathnerd15

## Homework Statement

13 equal charges q are placed at the corners of a polygon and as on a clockface, what is the net force on a test charge at the center Q? I assume the charge at the center is positive and the charge on the polygon so they repel

## Homework Equations

(1/4piEo)(qQ/r^2)

## The Attempt at a Solution

it seems there is a net force in one direction where there is a lack of symmetry in the Fx, Fy. for the situation of q=3 I calculated that there is a net charge upwared of
(1/4piEo)(qQ/r^2)(1-2sin(pi/6))=0
with the equilateral triangle pointing upward, sin 30 degrees=1/2

but as n sides of the polygon approaches infinity even in an odd polygon this would resemble a smooth circle and the charges would cancel out with Fnet=0? take out one charge in the 13 sided polygon and the charge motion and electrical force is in the direction of the missing charge?

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mathnerd15 said:

## Homework Statement

13 equal charges q are placed at the corners of a polygon and as on a clockface, what is the net force on a test charge at the center Q? I assume the charge at the center is positive and the charge on the polygon so they repel

## Homework Equations

(1/4piEo)(qQ/r^2)

## The Attempt at a Solution

it seems there is a net force in one direction where there is a lack of symmetry in the Fx, Fy. for the situation of q=3 I calculated that there is a net charge upwared of
(1/4piEo)(qQ/r^2)(1+r)
with the equilateral triangle pointing upward, sin 30 degrees=1/2

but as n sides of the polygon approaches infinity even in an odd polygon this would resemble a smooth circle and the charges would cancel out with Fnet=0? take out one charge in the 13 sided polygon and the charge motion and electrical force is in the direction of the missing charge?

You don't need all of that calculation. All you'd need is that the force towards the centre is the same for each of the charges situated on the polygon corners. You might as well call it 1. The the problem is reduced to resolved components of these equal forces, an almost pure geometrical or trigonometrical problem.

We know what you mean by not symmetrical but it is not true there is no symmetry.

Try it for 3 charges (disposed at 120° tp each other, say one at the bottom) and look at the horizontal sum of forces - easy - and then look at the vertical sum - not difficult. Then see if concolsions generalise.

mathnerd15 said:
with the equilateral triangle pointing upward, sin 30 degrees=1/2
Yes.

so for a circle with n=3, with the point at the bottom, Ey=(1-2sin(pi/6))=0 and Ex=0 since we rotated one point directly in -y direction

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mathnerd15 said:
so for a circle with n=3, with the point at the bottom, Ey=(1-2sin(pi/6))=0 and Ex=0 since we rotated one point directly in -y direction

OK and what is sin(π/6) ?

sin pi/6=sin 30degrees=1/2 so the y components go to zero...I'm not sure it's exactly 30 degrees when you divide the angle?

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Thought they might. It only remains to generalise.

And maybe you don't need to make a big meal of that either - maybe even the above is too complicated.

How about if what you did at the bottom node for the horizontal net force you did for each node in turn?

## 1. What is a polygon distribution?

A polygon distribution refers to a geometric arrangement of electric charges in a two-dimensional plane. It can also be referred to as a charge distribution in a polygon.

## 2. How does a polygon distribution affect the electric field?

A polygon distribution affects the electric field by creating areas of varying electric potential. The electric field lines will be closer together where there are more charges and farther apart where there are fewer charges.

## 3. What is the relationship between the shape of a polygon distribution and the electric field?

The shape of a polygon distribution will determine the direction and strength of the electric field. The electric field will be stronger in areas where the shape is more concentrated and weaker in areas where the shape is more spread out.

## 4. Can a polygon distribution have both positive and negative charges?

Yes, a polygon distribution can have both positive and negative charges. The overall distribution of charges will determine the net charge of the polygon.

## 5. How can the electric potential be calculated for a polygon distribution?

The electric potential can be calculated for a polygon distribution by using the formula V = kQ/r, where k is the Coulomb's constant, Q is the charge, and r is the distance from the charge to the point where the electric potential is being calculated. The electric potential at a point is the sum of the potentials due to each individual charge in the distribution.

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