Question

The three point charges +4.0 μC, -5.0 μC, and -9.0 μC are placed on the x-axis at the points x = 0 cm, x = 40 cm, and x = 120 cm, respectively. What is the x component of the electrostatic force on the -9.0 μC charge due to the other two charges? (k = 1/4πε0 = 9.0 × 109 N ∙ m2/C2)

Answer 1

0.4078 N towards right

Step-by-step explanation:

qo = 4 micro coulomb = 4 x 10^-6 C

qA = - 5 micro coulomb = - 5 x 10^-6 C

qB = - 9 micro coulomb = - 9 x 10^-6 Coulomb

OA = 40 cm = 0.4 m

OB = 120 cm = 1.2 m

AB = 120 - 40 = 80 cm = 0.8 m

Force on - 9 micro coulomb due to - 5 micro coulomb is

`F_(AB)=(Kq_(A)q_(B))/(0.8^(2)) =(9* 10^(9)* 5* 10^(-6)* 9* 10^(-6))/(0.8^(2))`

FAB = 0.6328 N rightwards

Force on - 9 micro coulomb due to 4 micro coulomb is

`F_(OB)=(Kq_(O)q_(B))/(1.2^(2)) =(9* 10^(9)* 4* 10^(-6)* 9* 10^(-6))/(1.2^(2))`

FOB = 0.225 leftwards

Net force = FAB - FOB = 0.6328 - 0.225 = 0.4078 N towards right

The force acting on 9 micro coulomb so the net force is also along X axis and directed rightwards.

Answer 2

Step-by-step explanation:

4μC will attract -9μC towards the centre and -5μC will repel it away from the centre. Both these forces are opposite to each other.

Force due to 4μC on -9μC towards the centre

= k x Q₁ Q₂/R² = 9 X 10⁹ X 4 X 10⁻⁶ X 9 X 10⁻⁶ / (1.2)² = 225 X 10⁻³ N/C

Force due to -5μC on -9μC away from the centre

= 9 x 10⁹ x 5 x 10⁻⁶x 9 x 10⁻⁶/( 0.8)² = 632.8 x 10⁻³ .N/C

Ner field =407.8 N/C.