Electrostatic force between two small spheres at separation 'r' is 'F'. Now both spheres are touched and again placed at separation 'r', then electrostatic force between them becomes (9 F /8). Then ratio of their initial charges is :-

Question

Electrostatic force between two small spheres at separation 'r' is 'F'. Now both spheres are touched and again placed at separation 'r', then electrostatic force between them becomes (9 F /8). Then ratio of their initial charges is :- (A) (3/4) (B) 2: 1 (C) (3/5) (D) (7/2)

Tardigrade Admin · Accepted Answer

F =( KQ 1 Q 2/ r 2) After touching (9 F /8)=( K ( Q 1+ Q 2)2/4 r 2) ⇒ (9/8)=((Q1+Q2)2/4 Q1 Q2) ⇒ (Q1/Q2)=(1/2)

Q. Electrostatic force between two small spheres at separation ' $r$ ' is ' $F$ '. Now both spheres are touched and again placed at separation ' $r$ ', then electrostatic force between them becomes $\frac{9 F}{8}$ . Then ratio of their initial charges is :-

$\frac{3}{4}$

$2 : 1$

$\frac{3}{5}$

$\frac{7}{2}$

$F = \frac{K Q _{1} Q _{2}}{r ^{2}}$
After touching $\frac{9 F}{8} = \frac{K ( Q _{1} + Q _{2} ) ^{2}}{4 r ^{2}}$
$\Rightarrow \frac{9}{8} = \frac{( Q _{1} + Q _{2} ) ^{2}}{4 Q _{1} Q _{2}}$
$\Rightarrow \frac{Q _{1}}{Q _{2}} = \frac{1}{2}$

Q. Electrostatic force between two small spheres at separation 'r' is 'F'. Now both spheres are touched and again placed at separation 'r', then electrostatic force between them becomes 89F​. Then ratio of their initial charges is :-

43​

2:1

53​

27​