# The Electric Field

Electric field and other phenomena for Charge inside cavity of a conductor

Tip: This isn't the place to ask a question because the teacher can't reply.

## Key Questions

• Physically, electric field represents the electric force per unit charge at each point in space. Mathematically, since force is a vector, electric field is a vector field, a mapping from each point in space to a vector value (a magnitude and direction).

Coulomb's law is an empirical formula first published by Charles Augustin de Coulomb in the 18th century. It describes the force between two point charges (charges that are small compared to the distance between them).

$\setminus \vec{F} = {k}_{e} \setminus \frac{{Q}_{1} {Q}_{2}}{{r}^{2}} \setminus \hat{r}$

where $F$ is the force of one charge on the other, ${Q}_{1}$ and ${Q}_{2}$ are the magnitudes of the two charges, $r$ is the distance between them, $\setminus \hat{r}$ is a unit vector pointing from one charge to the other. The quantity ${k}_{e} = 9 \setminus \times {10}^{9} N \setminus \cdot {m}^{2} {C}^{- 2}$ is a constant of proportionality. The force attractive if the charges are of opposite kind, and repulsive if they are of like kind.

This formula says the force between the two charges is directly proportional to the charge of each body, inversely proportional to the square of the distance between them, and directed along the line joining their centers.

From the point of view of either charge, therefore, the formula gives a both a magnitude and a direction, describing the force due to the other charge, at any given point in space relative to the other charge.

A magnitude and direction associated with each point in space is called a vector field. In this case that vector field, which describes the force per unit charge due to another charge, is called the electric field.

Suppose one of the charges (call it $Q$) is the one we really care about and the other charge (call it $q$) is just a hypothetical second charge whose sole purpose is to experience electric forces. The first one is called the source charge and the second one is called a test charge. Then Coulomb's law reads
$\setminus \vec{F} = {k}_{e} \setminus \frac{Q q}{{r}^{2}} \setminus \hat{r}$

Divide both sides by the magnitude of the test charge $q$ to find the electric field, which is the force per charge at any point in space:
$\setminus \vec{E} = \setminus \frac{\setminus \vec{F}}{q} = {k}_{e} \setminus \frac{Q}{{r}^{2}} \setminus \hat{r}$

The direction

#### Explanation:

Electric field is defined

• Electric field is a little bit more than a vector. It is a vector field, which is a function that maps each point in a volume to a vector. It associates a vector with each location in space.

This is because electric field represents the force per charge at each point in space, and force is a vector. Electric field, like force, has two aspects: a magnitude, and a direction.

So you could say that electric field is a vector (field) because force is a vector.

• The short answer is if they did cross, they would represent a location with two different strong electric field vectors, something that can't exist in nature.

Lines of force represent the strength of the electric field at any given point. Visually the denser we draw the lines, the stronger the field is.

Electric field lines reveal information about the direction (and the strength) of an electric field within a region of space. If the lines cross each other at a given location, then there must be two distinctly different values of electric field with their own individual direction at that given location. This could never be the case. Therefore the lines representing the field cannot cross each other at any given location in space.

Examples
If field lines were to cross what we need to do is merge them into a resultant field line. At each point in space we need to do vector addition on the field vectors from each source.

[Someone will have to confirm/correct whether it is correct to do vector addition with electric fields. If it is incorrect then we would perform vector addition on the force vectors instead and arrive at the same result.]

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