A charge of 1 μ C is divided into two parts such that their charges are in the ratio of 2:3 . These two charges are kept at a distance 1 m apart in vacuum. Then, the electric force between them (in N ) is -

Question

A charge of 1 μ C is divided into two parts such that their charges are in the ratio of 2:3 . These two charges are kept at a distance 1 m apart in vacuum. Then, the electric force between them (in N ) is - (A) 0.216 (B) 0.00216 (C) 0.0216 (D) 2.16

Tardigrade Admin · Accepted Answer

(q1/q2)=(2/3) q1 + q2 = 1Î¼C ⇒ q1 = (2/5)μ C , q2= (3/5)Î¼C F = (K q1 q2/r2) textF = (9 × 1 09 × (2/5) × (3/5) × 1 0- 1 2/(1)2) = 0.00216 textN

Q. A charge of $1 μ C$ is divided into two parts such that their charges are in the ratio of $2 : 3$ . These two charges are kept at a distance $1 m$ apart in vacuum. Then, the electric force between them (in $N$ ) is -

$0.216$

$0.00216$

$0.0216$

$2.16$

$\frac{q _{1}}{q _{2}} = \frac{2}{3}$
$q_{1} + q_{2} = 1 μ C$
$\Rightarrow$ $q_{1} = \frac{2}{5} μ C, q_{2} = \frac{3}{5} μ C$
$F = \frac{K q _{1} q _{2}}{r ^{2}}$
$F = \frac{9 \times 1 0 ^{9} \times \frac{2}{5} \times \frac{3}{5} \times 1 0 ^{- 12}}{( 1 ) ^{2}}$
$= 0.00216 N$

Q. A charge of 1μC is divided into two parts such that their charges are in the ratio of 2:3 . These two charges are kept at a distance 1m apart in vacuum. Then, the electric force between them (in N ) is -

0.216

0.00216

0.0216

2.16