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)

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.

f is prop to m1xm2 double m1 double force

a.   mass number from atomic number