Everything Science Grade 11

17.3 CHAPTER 17. ELECTROSTATICS

E = k

Q

r^2

=

(8. 99 × 109 )(5× 10 −^9 )

(0,3)^2

= 4, 99 × 102 N.C−^1

Example 6: Electric field 2

QUESTION

Two charges of Q 1 = +3nC and Q 2 =−4nC are separated by a distance of 50cm. What is

the electric field strength at a point that is 20cm from Q 1 and 50cm from Q 2? The point lies

between Q 1 and Q 2.

�

+3nC

x

10 cm 30 cm

�

-4nC

SOLUTION

Step 1 : Determine what is required

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

Step 2 : Determine what is given

We are given the magnitude of the charges andthe distances from the charges.

Step 3 : Determine how to approach the problem

We will use the equation:

E = k

Q

r^2

.

We need to work out the electric field for each charge separately and then add

them to get the resultant field.

Step 4 : Solve the problem

We first solve for Q 1 :

E = k

Q

r^2

=

(8. 99 × 109 )(3× 10 −^9 )

(0,2)^2

= 6, 74 × 102 N.C−^1