In free space, point charge C (1 C C, i.e. a nanocoulomb) is at and charge C is at .
[Note: Since the -coordinate of both charges is zero, the problem is two-dimensional in the plane as shown in Figure 35.]
- Find the field at the origin due to and .
- Where should a third charge C be placed in the plane so that the total field due to is zero at the origin?
Total field at the origin
= (field at origin due to charge at
) + (field at origin due to charge at
(The negative sign in front of the second term results from the fact that the direction from to is in the direction.)
Suppose the third charge
C is placed at
. The field at the origin due to the third charge is
where is the unit vector in the direction from to
If the position of the third charge is such that the total field at the origin is zero, then . There are two unknowns ( and ). We can write down two equations by considering the and directions.
Squaring and adding (3) and (4) gives
Substituting back from (5) into (1) and (2) gives and to 3 s.f.
Eight point charges of 1 nC each are located at the corners of a cube in free space which is 1 m on each side (see Figure 36). Calculate at
- the centre of the cube
- the centre of any face
- the centre of any edge.
- The field at the centre of the cube is zero because of the symmetrical distribution of the charges.
Because of the symmetrical nature of the problem it does not matter which face is chosen in order to find the magnitude of the field at the centre of a face. Suppose the chosen face has corners located at
then the centre (
) of this face can be seen from the diagram to be located at
The electric field at due to the charges at the corners and will then be zero since the field vectors due to equal charges located at opposite corners of the square cancel one another out. The field at is then due to the equal charges located at the remaining four corners of the cube, and we note from the symmetry of the cube, that the distance of each of these corners from will be the same. In particular the distance The electric field at due to the remaining charges can then be found using where to are the equal charges ( coulombs) and to are the vectors directed from the four corners, where the charges are located, towards . In this case since coulombs and for to we have
Suppose the chosen edge to be used connects
then the centre point
will be located at
By symmetry the field at due to the charges at and will be zero.
We note that the distances and are all equal. In the case of we calculate by Pythagoras that this distance is .
Similarly the distances and are equal to .
Using the result that gives
If where and are unit vectors in the and directions respectively, find the differential amount of work done in moving a C point charge a distance of 5 mm.
- From towards
- From towards
- The work done in moving a C charge through a distance of 5 mm towards is
- A similar calculation yields that the work done in moving the same charge through the same distance in the direction from to is