# Force at a point by continuous charge distribution....

In summary: With a bit of thought it is not too difficult to work out the distribution if it is not uniform, but it is still more effort than simply being told it.In summary, the question asks for the force on a test charge at the center of a ring of charge with a given charge distribution. The distribution is linear and non-uniform, but in a conductor the field inside is zero. However, this rule only applies to closed surfaces and not to a ring, so the distribution can be non-uniform. In general, it can be difficult to determine the charge distribution in conductors with imperfect symmetry.

## Homework Statement

This is more of a general question, but a simple example would be find the force on a test charge q at the center of a ring of charge with a total charge Q and a charge distribution given as λ(θ) =ksin(θ) where θ is measured clockwise with respect to the positive x-axis. The ring has radius R.

## Homework Equations

Coulomb's Law for continuous charge distributions.

## The Attempt at a Solution

Problems like this that I've seen often involve non-conducting materials (or it's not specified). My question is, what happens if we have a conducting material? It's not clear if you could set up the integration the same way or not.

In a conductor the field "inside' is zero because the charges are free to move and will naturally arrange themselves' in order to reach the lowest potential. All charge should be on the surface, and so there is no charge density inside the material. But I think I can reconcile this with the given charge density since it's linear, so I think we can say linearly there is a charge density.

## Homework Statement

This is more of a general question, but a simple example would be find the force on a test charge q at the center of a ring of charge with a total charge Q and a charge distribution given as λ(θ) =ksin(θ) where θ is measured clockwise with respect to the positive x-axis. The ring has radius R.

## Homework Equations

Coulomb's Law for continuous charge distributions.

## The Attempt at a Solution

Problems like this that I've seen often involve non-conducting materials (or it's not specified). My question is, what happens if we have a conducting material? It's not clear if you could set up the integration the same way or not.

In a conductor the field "inside' is zero because the charges are free to move and will naturally arrange themselves' in order to reach the lowest potential. All charge should be on the surface, and so there is no charge density inside the material. But I think I can reconcile this with the given charge density since it's linear, so I think we can say linearly there is a charge density.
The rule that there is no field inside a conductor only applies to closed surfaces, so not to a ring.
But if the ring is conducting, and the test charge is at the centre, then clearly the distribution would be uniform around the ring, not following the sort of distribution you described.

Here's an example of such a problem.

A conducting rod carrying a total charge of Q is bent into a semicircle of radius R centered at the origin. The charge density along the rod is given by
λ = λ0 sin(θ), where θ is measured clockwise from the +x axis. What is the magnitude of the electric force on a Q charged particle placed at the origin?

I agree the charge distribution should be uniform in a conductor. In a problem like this I'm not following how their could be variable charge density, since this would require that the system is dynamic correct?

I agree the charge distribution should be uniform in a conductor.
This is only true if symmetry demands a uniform distribution of charge. Circular rings or spherical shells are a classic examples of shapes that afford uniform charge distribution. Conducting objects with imperfect symmetry or corners or protrusions will not exhibit uniform charge density: Due to mutual repulsion, charge tends to flee to remote ends as much as possible.

In general it can be quite difficult to figure out the charge distribution in a conductor that lacks circular or spherical symmetry. In the question you quote, they have told you the distribution rather than require you to deduce it.

## 1. What is a continuous charge distribution?

A continuous charge distribution refers to a system where the electric charge is distributed continuously throughout a given space, rather than being localized or concentrated at specific points. This distribution can take various forms, such as a line, a surface, or a volume.

## 2. How is the force at a point determined by a continuous charge distribution?

The force at a point due to a continuous charge distribution is determined by integrating the electric force contributions from each element of the distribution. This integration takes into account the distance, direction, and magnitude of the force from each element, resulting in a net force at the desired point.

## 3. What is the formula for calculating the force at a point by a continuous charge distribution?

The formula for calculating the force at a point by a continuous charge distribution is F = ∫ (k * dq * q / r^2), where F is the net force, k is the Coulomb's constant, dq is the charge element, q is the charge of the element, and r is the distance between the element and the point of interest.

## 4. How does the distance from the point of interest affect the force by a continuous charge distribution?

The force by a continuous charge distribution is inversely proportional to the square of the distance from the point of interest. This means that as the distance increases, the force decreases, and as the distance decreases, the force increases.

## 5. Can the force at a point by a continuous charge distribution be both attractive and repulsive?

Yes, the force at a point by a continuous charge distribution can be both attractive and repulsive. This depends on the relative charges of the elements and the point of interest. If the charges are of the same sign, the force will be repulsive, and if the charges are of opposite signs, the force will be attractive.

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