**What is the electric field?**

He **electric field** It is the property that charged objects have to influence the space that surrounds them, which is perceived by other electrically charged bodies. But unlike the electric force between charges, the electric field depends only on the charge that produces it.

Michael Faraday (1791-1867), an English physicist, created the concept of a field by observing that any electric charge influences the space around it in such a way that it does not need to be in contact with another charge for interaction to occur.

It is not even necessary for the charges to be in a material medium, since the interaction can take place in a vacuum.

To visualize the shape of an electric field, suppose a point and positive charge, called +q, whose size is so small that it is not necessary to take its dimensions into account. The field that it produces is capable of affecting other charges, such as another positive test point charge qo.

The test charge is placed at different locations around +q, and since both are positive, the force that +q exerts on qo is repulsive.

Drawing the force vector on the charge qo at each point of the space it occupies, and removing it, we are left with a set of lines that emerge radially from the charge +q (see the image above, on the left).

By repeating the experience with a negative charge – q, the lines are also radial, but entering a – q. In both cases, the lines are tangent to the electric field vector of the charge, leaving it when it is positive, and entering it if it is negative.

**formula and units**

If there is an electric field in a region of space **AND, **an electric charge that experiences, thanks to it, a force given by:

**F **= what**AND**

So that:

The unit of the electric field in the SI System of units is newton/coulomb, which is abbreviated N/C. It is also common to express the electric field in terms of a scalar quantity called electric potential, in which case the unit for the field is the volt/meter (V/m).

**The electric field of a point charge**

Field **AND **is produced by some object with charge q. By making the test charge very small, that is, by letting qo tend to 0, the vector **AND** is:

With **F**or the force between q and qo.

The intention in taking the limit is to make the test charge small enough so that its field does not alter the one that is to be calculated.

If q is a point charge, according to Coulomb’s law, the force between charges q and qo, both separated by a distance r, is given by:

In this equation, k is the electrostatic constant and the unit vector in the direction of the line joining q and qo is:

Substituting this expression into the field definition, you get:

So the field **AND**produced by the point charge q at point P, is:

Thus, **AND** it does not depend on the test load, but on the load that produces it. The *magnitude *of the field is directly proportional to the magnitude of the charge, and inversely proportional to the square of the distance between the charge and the point P.

And as said at the beginning, the *address* of the field is radial and the direction is outgoing to the charge when it is positive, and incoming when it is negative.

**Electric field strength**

The electric field is vectorial, and its intensity refers to its module or magnitude, which is denoted without bold. For a point charge, the intensity of its electric field is simply:

And because it is the module of a quantity, it is always positive.

For example, the intensity of the electric field produced by a charge q = − 4.3 μC (μC is read “microcoulomb” and equals one millionth of a coulomb), at a distance of 2 cm from the charge, is:

Note that the distance of 2 cm was converted to meters by multiplying by the power of 10−2, since the electrostatic constant is in SI units. And although the charge is negative, the intensity of the field it produces is always positive, but the electric field vector is entering the charge, as previously explained.

**Electric Field Examples**

**1. Electric field of a discrete distribution of charges**

A set of point charges is called *discrete charge distribution*. In that case, the resulting electric field at point P is calculated by applying the *superposition principle*which is the vector sum of the field that each of the charges produces at P:

**AND**net = **AND**1 + **AND**2 + **AND**3 + …

The following image shows a distribution composed of five point charges and the electric field that each one produces at point P:

The charges q3 and q5 are negative and the field they produce is entering them. They are distinguished in blue. For their part, the charges q1, q2 and q4 are positive, creating an outgoing field in red.

**2. Electric field of a continuous distribution of charges**

A continuous charge distribution consists of an extended, electrically charged object, like the one shown in the following figure. Given that the object has appreciable dimensions, the field that a part of the body produces in P is significantly different from that produced by another part that is further away (or closer) from P.

Suppose a small electric charge is taken from said object, called dq and assumed to be positive, which produces a small contribution to the total electric field at P. This contribution is a differential of the electric field vector d**AND**.

Since the charge dq is very small, its field is like that of a point charge, so the equation seen before can be applied:

To obtain the total field of the object at point P, the contributions of all the dq that can be taken on the object are added. This leads to the integral:

**solved exercise**

A point charge Q = 2.0 × 10−8 C is placed at a point P within an electric field, in which it experiences an upward force of magnitude 4.0 × 10−6 N. Calculate:

a) The electric field at P

b) The force on a charge q = −1.0 × 10−8 C located at P.

**Solution to**

Let E be the magnitude of the electric field in which the charge Q is placed. By virtue of this field, the charge experiences the upward force of magnitude F, so that:

F = Q∙E

So:

E = F /Q = 4.0×10–6 N/ 2.0×10–8 C = 200 N/C.

Being positive the charge, the force and the field have the same direction and sense.

**solution b**

The magnitude of the force acting on q is:

Since this charge is negative, force and field have the same direction, but opposite directions.

**References**

Bauer, W. 2011. Physics for Engineering and Science. Volume 2. Mc Graw Hill.

Electric field and potential of a point charge. Retrieved from: sc.ehu.es.

Resnick, R. (1999). Physical. Vol.1.3to ed. in Spanish. Continental Publishing Company SA de CV

Sears, Z. (2016). University Physics with Modern Physics. 14th. Ed. Volume 1. Pearson.

University Physics. Electric Field. Vol. 2. Retrieved from: openstax.org.