Everything Science Grade 11

17.2 CHAPTER 17. ELECTROSTATICS

because force is a vector quantity.

Step 3 : Determine what is given

We are given all the charges and all the distances.

Step 4 : Calculate the forces.

Force of Q 1 on Q 2 :

F = k

Q 1 Q 2

r^2

= (8, 99 × 109 )

(2× 10 −^9 )(1× 10 −^9 )

(2× 10 −^4 )

= 4, 5 × 10 −^5 N

Force of Q 3 on Q 2 :

F = k

Q 2 Q 3

r^2

= (8, 99 × 109 )

(1× 10 −^9 )(3× 10 −^9 )

(4× 10 −^4

= 1, 69 × 10 −^5 N

Both forces act in the same direction because the force between Q 1 and Q 2 is

repulsive (like charges) and the force between Q 2 and Q 3 is attractive (unlike

charges).

Therefore,

Ftot = 4, 50 × 10 −^5 + 4, 50 × 10 −^5

= 6, 19 × 10 −^5 N

We mentioned in Chapter ?? that charge placed on aspherical conductor spreads evenly along the
surface. As a result, if we are far enough from the charged sphere, electrostatically, it behaves asa
point-like charge. Thuswe can treat spherical conductors (e.g. metallicballs) as point-like charges,
with all the charge acting at the centre.

Example 4: Coulomb’s Law: challenging question

QUESTION

In the picture below, X is a small negatively charged sphere with a massof 10kg. It is

suspended from the roof by an insulating rope which makes an angle of 60 ◦with the roof. Y

is a small positively charged sphere which has the same magnitude of charge as X. Y is fixed

to the wall by means ofan insulating bracket. Assuming the system is inequilibrium, what is

the magnitude of the charge on X?