Everything Science Grade 11

CHAPTER 17. ELECTROSTATICS 17.3

The magnitude of the electric field at a point asthe force per unit charge. Therefore,

E =

F

q

E and F are vectors. Fromthis we see that the force on a charge q is simply:

F = E· q

The force between two electric charges is given by:

F = k

Qq

r^2

.

(if we make the one charge Q and the other q.) Therefore, the electricfield can be written as:

E = k

Q

r^2

The electric field is the force per unit of charge and hence has units of newtons per coulomb.

As with Coulomb’s lawcalculations, do not substitute the sign of the charge into the equation for
electric field. Instead, choose a positive direction, and then either add orsubtract the contributionto
the electric field due toeach charge dependingupon whether it points in the positive or negative
direction, respectively. See video: VPlno at http://www.everythingscience.co.za

See simulation: VPlnv at http://www.everythingscience.co.za)

Example 5: Electric field 1

QUESTION

Calculate the electric field strength 30cm from a 5nC charge.

�

+5nC

x

30 cm

SOLUTION

Step 1 : Determine what is required

We need to calculate the electric field a distance from a given charge.

Step 2 : Determine what is given

We are given the magnitude of the charge and the distance from the charge.

Step 3 : Determine how to approach the problem

We will use the equation:

E = k

Q

r^2

.

Step 4 : Solve the problem