The electrostatic force on a small sphere of charge 0.4 μ C due to another small sphere of charge -0.8 μ C in air is 0.2 N. The distance between the two spheres is

Question

The electrostatic force on a small sphere of charge 0.4 μ C due to another small sphere of charge -0.8 μ C in air is 0.2 N. The distance between the two spheres is (A) 6 cm (B) 8 cm (C) 10 cm (D) 12 cm

Tardigrade Admin · Accepted Answer

F=(q1 q2/4 π ε0 r2) ⇒ r2=(q1 q2/4 π ε0 ⋅ F)=((9 × 109)(0.4 × 10-6)(0.8 × 10-6)/0.2) ∴ r2=144 × 10-4 ⇒ r=12 × 10-2 m =12 cm

Q. The electrostatic force on a small sphere of charge $0.4 μ C$ due to another small sphere of charge $- 0.8 μ C$ in air is $0.2 N$ . The distance between the two spheres is

6 cm

8 cm

10 cm

12 cm

$F = \frac{q _{1} q _{2}}{4 π ε _{0} r ^{2}}$
$\Rightarrow r^{2} = \frac{q _{1} q _{2}}{4 π ε _{0} \cdot F} = \frac{( 9 \times 1 0 ^{9} ) ( 0.4 \times 1 0 ^{- 6} ) ( 0.8 \times 1 0 ^{- 6} )}{0.2}$
$∴ r^{2} = 144 \times 1 0^{- 4} \Rightarrow r = 12 \times 1 0^{- 2} m = 12 c m$

Q. The electrostatic force on a small sphere of charge 0.4μC due to another small sphere of charge −0.8μC in air is 0.2N. The distance between the two spheres is