The factor simplifies later formulae. If and are of the same sign then the force is positive and so is repulsive. If they are of opposite signs the force is negative and so is attractive. Calculating the magnitude of the coefficients we find that

so that the

If a unit charge is placed at a point , the force it experiences is
called the *electric field* at . A *point
charge*, , at produces at an *electric field*
(i.e. the force a unit charge feels there),

which is satisfied by setting

is called the electrostatic potential. For the point charge situation described above this is

Note that :

- The potential due to several charges is obtained by adding the
individual potentials together so that

- In general, surfaces of constant are perpendicular to the electric field .

__Example 3. .1__Consider an electron of mass and charge moving in a circlar
orbit of radius
about a fixed charge at the origin . Obtain the speed of
the electon.

The electron feels a
centrifugal force of , due to the circular motion and this
is
balanced by the attractive Coulomb force of
. Therefore,

Hence, the electron's speed is

__Example 3. .2__Two positive charges of strength are located at on
the z-axis. Find the electric field on the z-axis and sketch the
field lines in the x-z plane.

The position vector of a general point in the plane is

The situation is shown in Figure 3.1. On the z-axis (), we need to consider the direction of the electric field due to each of the point charges. Starting to the left of , both point charges give a component of the electric field that is in the negative direction. Between the point charges the left hand charge gives a field component that points to the right and the right hand charge points to the left. Finally, to the right of the charges, both give electric field components pointing to the right. Thus, we have

To calculate the electric field at a general point we need to work out the distance from each point charge and the unit vector. For example,

and

The unit vectors are

Using these results, the electric field is given by

The resulting electric field is shown in Figure 3.2

__Example 3. .3__Charges and are located on the x-axis at (-a,0,0) and
(a,0,0) respectively. (i) Show that there is one neutral point, ,
where the electric field vanishes, i.e.

(ii) Draw a rough sketch of the electric field lines.

(i) Consider the general point, , . The situation is shown in Figure 3.3.

The electric fields due to each of the point charges are only colinear (and therefore can cancel at a neutral point) when is on the x-axis. Here they cancel at a point . Consider the electric field on the x-axis. As in the previous example, the form of the electric field depends on where the point is. Thus,Thus, there is only a possible neutral point in the first and third regions. To obtain the position of the neutral point we solve

Rearranging we obtain

Only the first of these roots lies in the correct range for . The second root lies in the interval where it is clear that there can be no neutral point. Thus, the only neutral point is at .

(ii) To plot the field lines we can use the electrostatic potential
where
. We can verify that

The electric field lines are given by contours of constant . Thus, we obtain the field lines as shown in Figure 3.4. Half the flux from the point source goes into and half goes off to infinity.