- Coulomb's law: F = k q
_{1}q_{2}/ r² - Electric field: E = k q / r²
- Both the electric force and electric field are vector quantities.
- The above relations are valid for point charges or spherical charges.
- For a collection of charges, the total force or field is given by summing the contributions from each charge individually -- the superposition principle.

We can graphically represent the electric field at all points in space by drawing electric field lines. A drawing of the electric field lines immediately show several things.

- The electric field vector,
**E**, is tangent to the electric field lines at each point. - The magnitude of the electric field is proportional to the density of field lines near a particular location.

The electric field lines of an isolated positive charge point radially outward. For an isolated negative charge the lines point radially inward. (Draw examples.) For more complicated arrangements of charge, the lines are normally curved, not straight.

There are some rules to follow for drawing electric field lines:

- The lines must begin on positive charges (or at infinity), and end on negative charges (or at infinity).
- The number of lines drawn leaving a positive charge or entering a negative charge is proportional to the magnitude of the charge.
- No fields lines can cross.

The ability of charges to move freely in a conductor results in some special properties.

- The electric field is zero everywhere inside the conductor.
- Any excess charge on an isolated conductor resides entirely on its surface.
- The electric field just outside a charged conductor is perpendicular to the conductor's surface.
- On an irregularly shaped conductor, the charge tends to accumulate at locations where the radius of curvature of the surface is smallest, that is, at sharp points.

The first property is true since if the electric field is not zero inside a conductor, then some of the sea of mobile charges inside would move until the electric field became zero.

The second property is a result of the 1/r² behavior of the electrical force.

The third property becomes evident after considering what happens if it is not true. If the field just outside the conductor has a component parallel to the surface of the conductor, then it will produce a force on surface charges that is parallel to the surface. These charges are free to move along the surface of the conductor (but not off the conductor), and so they will, until the parallel component of the force vanishes. It is important to recall that the electric field not only acts on the surface charges, it is produced by the surface charges. The movement of the charges changes the electric field until equilibrium is established.

The fourth property is harder to prove precisely, but is the reason why it can be dangerous to be on a hilltop or mountaintop during a thunderstorm. In such locations, a standing person can become a "sharp" point, resulting in a concentration of electric field lines at the person, and making him/her a target for a lightning strike.

I will skip sections 15.8 and 15.9.

From section 10 we will be discussing electric flux. Here we will learn only how flux is calculated, and won't make further use of it. The primary reason to discuss this now is to learn about how flux is calculated to prepare for when we meet it again in the case of magnetic fields.

Consider a region of uniform electric field, and a surface of area A perpendicular to the field. We define the flux as the amount of electric field passing through this area:

F = E A

where E is the value of the electric field in N/C and A is the area in m².
The flux has units of Nm²/C.
If the surface is not perpendicular to the field then the flux is given by:
F = E A cosq

where q is the angle between the perpendicular to the surface and the electric field.
A 40cm diameter loop is rotated in a uniform electric field until the position of maximum electric flux is found.
The flux in this position is measured to be 5.2×10^{5}Nm²/C.
Calculate the electric field strength in this region.

Begin by drawing a sketch of the uniform electric field, and a circular loop.
The loop forms the boundary of a circular surface of area A = π(0.20m)² = 0.126m².
Using the definition for flux, we have F = E A cosq.
Since E and A are fixed, the flux is maximum when cosq = 1.
This occurs when the loop is perpendicular to the field.
Now we can solve for the electric field.

E = F/A = (5.2×10^{5}Nm²/C)/0.126m² = 41×10^{5} = 4.1×10^{6}.

Calculate the flux through half a spherical shell of radius 10cm due to a charge of -4×10^{-9}C located at the center of the shell.

The electric field from a point charge is constant on a spherical shell centered on the charge.
Also, the electric field points radially outward; it is perpendicular to the surface of the spherical shell at every point where it penetrates the shell.
Therefore, even though the surface is not flat, we can still calculate the flux as the product of the field times the area. The area of half a spherical shell is A = 2πr². The electric field is E = k q / r².
The flux is

F = E A = (k q / r²)(2πr²) = 2πkq = 2π(9×10^{9}Nm²/C²)(-4×10^{-9}C) = -226Nm²/C.