There are only two charges, positive and negative.

The unit of charge is the coulomb (C). The charge of an electron (or a proton) is e = 1.60 x 10-19 C.

The smallest amount of charge that has been found is e, so any charge of magnitude q is an integer multiple of e. q = Ne, where N is an integer.

Ex. 1 - How many electrons are there in one coulomb of negative charge?

Friction can transfer electric charge, usually by moving electrons.

Law of conservation of electric charge - During any process, the net electric charge of an isolated system remains constant.

Like charges repel, unlike charges attract.Substances that readily conduct electric charge are called electrical conductors.

Materials that conduct electric charge poorly are called electrical insulators.

When one object is given a net electric charge by placing it in contact with a charged object it is called charging by contact.

If a charged object is held close to another object, and the second object is temporarily grounded before the first object is removed; the second object is left charged opposite the initial charge.

This is charging by induction.

Charging by contact makes both objects the same charge. Charging by induction will make the two objects opposite charges.

When point charges q1 and q2 are separated by distance r, the force between them is calculated by Coulomb’s Law.

F = k•q1•q2/ r2

k is a proportionality constant whose value is k = 8.99 x 109 N•m2/C2

The force is attractive if the charges are opposite, and repulsive if the charges are the same.

k is often expressed in terms of ε0, by writing k = 1/(4πε0).

ε0 is called the permittivity of free space and has a value of ε0 = 1/(4πk) = 8.85 x 10-12 C2/(N•m2).

Ex. 2 - Two objects, whose charges are +1.0 and -1.0 C, are separated by 1.0 km. Find the magnitude of the attractive force that either charge exerts on the other.

Ex. 3 - In the Bohr model of the hydrogen atom, the electron (-e) is in orbit about the nuclear proton (+e) at a radius of r = 5.29 x 10-11 m. Determine the speed of the electron assuming the orbit to be circular.

When more than two point charges are involved, the force on any one charge is the vector sum of the forces due to each individual charge.

Ex. 4 - Figure 18.14a shows three point charges that lie along the x axis in a vacuum. Determine the magnitude and direction of the net electrostatic force on q1 .

Ex. 5 - Figure 18.15a shows three point charges that lie in the x,y plane in a vacuum. Find the magnitude and direction of the net electrostatic force on q1.

A small point charge, called a test charge, may be used to determine the extent to which the surrounding charges generate a force.

Ex. 6 - The positive test charge in Figure 18.16 is q0 = +3.0 x 10-8 C and experiences a force F = 6.0 x 10-8 N in the direction shown in the drawing. (a) Find the force per coulomb that the test charge experiences. (b) Using the result of part (a), predict the force that a charge of +12 x 10-8 C would experience if it replaced q0.

The electric field E that exists at a point is the electrostatic force F experienced by a small test charge q0 placed at that point divided by the charge itself:

E = F/q0

The electric field is a vector, and its direction is the direction of of the force on the charge.The unit is the newton per coulomb (N/C).

The surrounding charges are what creates an electric field at a given point.

Ex. 7 - In figure 18.18a the charges on the two metal spheres and the ebonite rod create an electric field E at the spot indicated. The field has a magnitude of 2.0 N/C and is directed as in the drawing. Determine the force on a charge placed at that spot, if the charge has a value of (a) q0 = +18 x 10-8 C and

(b) q0 = -24 x 10-8 C.

Multiple electric fields add in the same way as vectors. Vector addition can be used to find the net field at a particular point.

Ex. 9 - There is an isolated charge of q = +15 µC in a vacuum. Using a test charge of q0 = +0.80 µC, determine the electric field at point P, which is 0.20 m away.

From Coulomb’s law, the force exerted on a test charge q0 by a charge q is F = k•q•q0/ r2. Since E = F/q0, E is also equal to k•q•q0/ r2 divided by q0. q0 cancels out, and we are left with E = k•q/ r2.

E = k•q/ r2 shows that the electric field does not depend on the test charge.

E = k•q/ r2 If q is positive, then E is directed away from q. If q is negative, E is directed toward q, since a negative charge attracts a positive charge.

Ex. 9 - Two positive point charges, q1 = +16 µC and q2 = +4.0 µC, are separated in a vacuum by a distance of 3.0 m. Find the spot on the line between the charges where the net electric field is zero.

A parallel plate capacitor has two plates of different charge with a space between them. The charges are distributed uniformly over each plate.

The electric field points from the positive plate to the negative and is perpendicular to both.

Using Gauss’ law, the electric field has a magnitude of: E = q/ ε0A.E = q/ ε0A = σ/ ε0

σ denotes charge per unit area (σ = q/A) and is called charge density.

In a parallel plate capacitor, the field has the same value at all places between the plates. The field does not depend on the distance from the charges, as it does in a field created by an isolated point charge.