Two particles with charges (+3.72 C ) and (+1.86 C ) are some distance apart. If (20 % ) of the charge is transferred from first particle to second particle then the electrostatic force between them is

Step-by-step solution

Given, charge on the first particle, (Q₁=+3.72 C ) and charge on second particle, (Q₂=1.86 C ) Then the electrostatic force between charges, ( aligned F₁ & = k Q₁ Q₂ R^2 & = k R^2 (3.72 1.86) 10⁻¹² ~N F₁ & = k R^2 (6.9192 10⁻¹² ) aligned ) If (20 % ) of (Q₁ ) is given to (Q₂ ), ( aligned & Q₂^ =Q₂+Q₁ 20 100 =1.86+3.72 0.20 & Q₂^ =2 604 C & and Q₁^ =Q₁ 80 100 =2.976 C aligned ) Hence, ( aligned & F₂= k R^2 Q₁^ Q₂^ = k R^2 2.976 2.604 10⁻¹² & F₂= k R^2 7.7495 aligned ) So, % increment in (F₂ ), ( aligned F₂-F₁ F₁ 100