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# Electrostatics Study Guide (page 2)

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Updated on Sep 27, 2011

#### Example

Consider two objects, one made of rubber and the other of nylon that is electrically neutral. You rub the two objects together and the rubber becomes charged with –2.88 · 10–16 Coulombs. What is the nylon's charge, and how many electrons have been shifted to the rubber from the nylon in the process?

#### Solution

Initially, both objects have a zero net charge, and, because the system will be considered to be isolated, the same amount of charge will be retrieved in the final state.

qrubber = –2.88 · 10–16 C

qnylon = ?

N =?

qrubber + qnylon = 0 C

–2.88 · 10–16 C + qnylon = 0 C

qnylon = – (–2.88 · 10–16 C)

qnylon = + 2.88 · 10–16 C

The number of electrons that make up the charge of the rubber is:

N = qrubber/qe

N = (–2.88 · 10–16 C)/(–1.6 · 10–19 C) = 1,800 electrons pass from nylon to rubber.

## Electric Forces and Coulomb's Law

We have talked about interaction between materials in the electrostatic series. We can summarize that there are two types of electrical interactions: attraction and repulsion. Attraction forces are established between particles of different charge sign (positive and negative charges). Repulsion forces act between charges of the same kind.

The quantitative expression of the electrical force is known as Coulomb's law (from Charles de Coulomb, 1736–1806). If one considers the charges to be pointlike (that is, all charges gather in one point), then the law says: The magnitude of the electrical force exerted by a point-like charge q1 on a charge q2 is proportional to the product of the charges and inversely proportional to the square of the distance between the charges.

Where the proportionality constant is:

k = 9 · 109N · m2/C2

The electric force, as are all other forces, is a vector. The direction is given by the type of particles interacting, as shown in Figures 13.2 and 13.3 and defined previously.

Although the names given to the forces are different to show the action charge and the test-charge (for example, F21 indicates that charge 2 is acting on charge 1), the absolute value is the same.

F21 = F12

Also shown in the figures is the opposite direction of interaction.

In a vector form, the law is expressed as:

· r

One can see that the new formula shows that the direction of the electric force is the same as the direction between the two charges.

### Coulomb's Law

The magnitude of the electrical force exerted by a point-like charge q1 on a charge q2 is proportional to the product of the charges and inversely proportional to the square of the distance between the charges.

#### Example 1

Two amber beads are brought in proximity after being charged with 10 and 50 electrons. They are placed at a distance of 20 cm from each other. Find the force acting on each bead.

#### Solution 1

First, convert to the units necessary to work with Coulomb's law and then find the charges. Then, calculate the forces on each object.

R = 20cm = 0.2m

q1 = 10 · (–1.6 · 10–19 C) = –1.6 · 10–18 C

q2 = 50 · (–1.6 · 10–19 C) = –8.0 · 10–18 C

F21 = F12 = ?

F21 = = 9 · 10–9 N · m2/C2 ·

F21 = 9 · 10–9 N · m2/C2 ·

F21 = 28.8 · 10–23N

F21 = F12 = 28.8 ·10–23N

The charges are the same sign, and therefore, the force is of repulsion.

If there are three or more objects charged and interacting, you will have to find the force that one charge is acted upon by all the other neighboring charges, and then find the net force. Remember that the force will be a vector, so the net force is the sum of vectors (value and direction).

#### Example 2

Consider three equal charges placed on the x-axis such that the distances between are 100 cm and 300 cm between charge 1 and charge 2 and between charge 1 and charge 3, respectively. Find whether the charge in the middle is at rest or not.

#### Solution 2

In order to find our answer, we start by converting all values to SI. Then, we will find the net force on the middle charge and see if it is zero or nonzero. In the case the force is zero, the charge is at rest (Newton's second law). If the net force is nonzero, the charge will be accelerated in the direction of the net force.

r12 = 100cm = 1m

r13 = 300cm = 3m

r23 =300 cm – 100 cm = 2 m = 2 · r12

q1 = q2 = q3 = q

Fnet = ?

Charge 2 is acted upon by two forces because of repulsion with the charges 1 and 3. The forces are F23 exerted by charge 3 on charge 2 and F21 exerted by charge 1 on charge 2. As you can see from the figure, they are in opposite directions.

If we consider the positive direction as in the figure, then F21 is positive and F23 is negative.

F21 =

F23 =

As mentioned in the problem statement:

r23 = 2 · r12 = 2 · r21

The last term in the equality is possible to write because we are interested in the absolute value and not the direction (which was considered in drawing the figure).

F21 =

Fnet = F23 + F21

Fnet = –F23 + F21

This result tells us that the charge is acted upon by a net force and subsequently will be accelerated by an acceleration that can be calculated based on Newton's second law if the mass of the charge is known:

Fnet = m · a

The force is in the direction of F21 , meaning the object is accelerated toward the positive x direction.

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