Electrostatics - AP Physics 1

Two principal realizations help with solving this problem, both derived from Gauss’ law for electricity:

1. The excess charge on an ideal conducting sphere is uniformly distributed over its surface

2. A uniform shell of charge acts, in terms of electric force, as if all the charge were contained in a point charge at the sphere’s center

With these realizations, an application of Coulomb’s law answers the question. If is the point charge outside the sphere, then the force on is:

In this equation, is Coulomb’s constant, is the excess charge on the spherical conductor, and is total distance in meters of from the center of the conducting sphere.

Using the given values in this equation, we can calculate the generated force:

This question is very simple if you realize that the force experienced by both charges is equal.

The definition of the two electrostatic forces are given by Coulomb's law:

In this question, we can rewrite this equation in terms of our given system.

It doesn’t matter if the charges of the two particles are different; both particles experience the same force because the charges of both particles are accounted for in the electrostatic force equation (Coulomb's law). This conclusion can also be made by considering Newton's third law: the force of the first particle on the second will be equal and opposite the force of the second particle on the first.

Since the forces are equal, their ratio will be .

Coulomb's law gives the relationship between the force of an electric field and the distance between two charges:

The strength of the force will be inversely proportional to the square of the distance between the charges.

When the distance between the charges is doubled, the total force will be divided by four (quartered).

The net force on the electron is the sum of the forces between the electron and each of the protons:

These forces are given by Coulomb's law:

Using the numbers given, we get:

Because opposite charges attract, points left (the negative direction) and points right (the positive direction).

Therefore, the net force is

Because this value is positive, the direction is rightward.

Use Coulomb's Law

Plug in known values and solve.

A negative value for electric force indicates an attractive force. This makes sense since our two charges have opposite signs. Since we're asked for magnitude, all answer choices are positive.

Use Coulomb's law.

Plug in known values and solve.

Note that the force between two charges of the same sign (both positive or both negative) is positive. This indicates the force is repulsive, which makes sense since both charges are positive.

Use Coulomb's law.

Plug in known values and solve.

Note that this force is positive, which means it's repulsive.

Use Coulomb's law.

Plug in known values and solve.

Note that this force is positive, which means that it's repulsive.

The force of attraction/repulsion between two point charges is given by Coulomb's Law:

If the charges are of like sign, then there well be a repulsive force between the two. Alternatively, if the net force is positive, it is repulsive; if it is negative, it is attractive.

Therefore, the force of repulsion between the two charges is:

The force of attraction/repulsion between two point charges is given by Coulomb's Law:

If the charges are of like sign, then there well be a repulsive force between the two.

Work is given as the dot product of force and distance. However, in this case, force is also dependent on distance.

The amount of work required to move a charge an incremental distance, , is given as:

The negative sign in this case is to account for repulsion.

The total work to change distances between charges can then be found by taking the integral with respect to distance:

Since are constants, they can be factored out of the integral:

Coulomb's law in vector notation is given as:

, where is Coulomb's constant, and are the two charges, is the distance between the charges squared, and is the unit vector going from one charge to another.

To write this in vector notation, we have to know the unit vector going from the negative to the positive charge, since we're trying to determine the force on the positive charge. Since they are both sitting on the -axis, with the negative charge to the left of the positive, the unit vector will be going in the direction of positive :

We know that

We know that , , and . Putting this together:

We can rewrite this as:

To determine this, we have to solve for in Coulomb's law and then determine its constants.

Recall that the magnitude of the electrostatic force between point charges is given as:

, is the force given in , and are the charges given in and is distance given in

Solving for ,

Writing out the terms on the left in their units:

Therefore, is given in

Coulomb's law shows that the force between two charged particles is inversely proportional to the square of the distance between the particles.

If the distance between the charges is reduced by , that means the is squared in the denominator and the will flip up to the top to give time the original force. More explicitly, if we plug in the given information the initial force will be:

Use Coulomb's law:

, where is Coulomb's constant, are charges of the two points and is the distance between the charges.

This is a question where knowing how to effectively sift through a problem statement and choose only the information you need will really help. We are given a bunch of values, but only need to know one thing, which is that the distance between the two charges is doubled.

Coulomb's law is as follows:

We can rewrite this for the initial and final scenarios:

We can divide one equation by the other to set up a ratio:

We know that the final radius is double the intial, which is written as:

Substituting this in we get:

Rerranging for the final force, we get:

First lets set up two axes. Have be to the right of charge 3 and 2 in the diagram and be above charges 1 and 2 in the diagram with charge 2 at the origin.

Coloumb's law tells us the force between point charges is

The net force on charge 2 can be determined by combining the force on charge 2 due to charge 1 and the force on charge 2 due to charge 3.

Since charge 1 and charge 2 are of opposite polarities, they have an attractive force; therefore, charge 2 experiences a force towards charge 1 (in the direction). By using Coloumb's law, we can determine this force to be

in the direction

Since charge 2 and 3 have the same polarities, they have a repulsive force; therefore, charge 2 experiences a force away from charge 2 (in the direction). By using Coloumb's law, we can determine this force to be:

in the -direction

If we draw out these two forces tip to tail, we can construct the net force:

From this, we can see that and create a right triangle with the net force on charge 2 as the hypotenuse. By using the Pythagorean theorem, we can calculate the magnitude of the net force:

Use Coulomb's law.

Plug in known values and solve.

Note that a positive value for electric force corresponds to a repulsive force. This should make sense since the charge on both particles are the same sign (positive).

Use Coulomb's law.

Note that the electric force between two charges of the same sign (both positive or both negative) is a positive value. This indicates a repulsive force.

Use Coulomb's law to find the electric force between the charges: