Question

The electrostatic force on a small sphere of charge 0.4 μC due to another small sphere of charge – 0.8 μC in air is 0.2 N.

(a) What is the distance between the two spheres?

(b) What is the force on the second sphere due to the first?

Answer 1

(a) From the formula

F = (Q1 Q2)/(4πεo r2 )

Substituting the values, we get

r = 0.12 m = 12 cm.

(b) Force on second sphere due to first

F21 = (Q1 Q2)/(4πεo r2 )

This comes to be 0.2 N which means force on second sphere due to first. It confirms Newton’s third law.

Answer 2

Answer:

[a ] Let charge on small sphere is denoted by q₁

Charge on large sphere is denoted by q₂

force is denoted by F

Given, q₁ = 0.4μC = 0.4 × 10⁻⁶ C

q₂ = 0 8μC = 0.8 × 10⁻⁶ C

F = 0.2 N

∴ use Coulombs law,

F = Kq₁q₂/r²

0.2 = 9 × 10⁹ × 0.4 × 10⁻⁶ × 0.8 × 10⁻⁶/r²

r² = (9 × 0.32) × 10⁻³/0.2

r² = 14.4 × 10⁻³

r² = 144 × 10⁻⁴

Taking square root both sides,

r = 12 × 10⁻² m = 12cm

[b ] According to Newton's third law,

=> |F₁₂| = |F₂₁|

so, F₁₂ = F₂₁ = 0.2N { attractive nature.}