Total charge Q is broken in two parts Q1 Q2 and they are placed at a distance R from each other. The maximum force of repulsion between them will occur when

Question

Total charge Q is broken in two parts Q1 Q2 and they are placed at a distance R from each other. The maximum force of repulsion between them will occur when (A) Q2=(Q/R), Q1=Q-(Q/R) (B) Q2=(Q/4), Q1=Q-(2 Q/R) (C) Q2=(Q/4), Q1=Q-(3 Q/4) (D) Q2=(Q/2), Q1=Q-(Q/2)

Tardigrade Admin · Accepted Answer

Q1+Q2=Q (i) Q2=Q-Q1 (ii) F =( KQ 1( Q - Q 1)/ R 2) (iii) For F to be maximum (d F/d Q1)=0 (1/ R 2)[ KQ -2 KQ 1]=0 ⇒ Q 1=( Q /2) Q2=Q-(Q/2)=(Q/2)

Q. Total charge $Q$ is broken in two parts $Q_{1} & Q_{2}$ and they are placed at a distance $R$ from each other. The maximum force of repulsion between them will occur when

$Q_{2} = \frac{Q}{R}, Q_{1} = Q - \frac{Q}{R}$

$Q_{2} = \frac{Q}{4}, Q_{1} = Q - \frac{2 Q}{R}$

$Q_{2} = \frac{Q}{4}, Q_{1} = Q - \frac{3 Q}{4}$

$Q_{2} = \frac{Q}{2}, Q_{1} = Q - \frac{Q}{2}$

$Q_{1} + Q_{2} = Q$ (i)
$Q_{2} = Q - Q_{1}$ (ii)
$F = \frac{K Q _{1} ( Q - Q _{1} )}{R ^{2}}$ (iii)
For $F$ to be maximum $\frac{d F}{d Q _{1}} = 0$
$\frac{1}{R ^{2}} [K Q - 2 K Q_{1}] = 0$
$\Rightarrow Q_{1} = \frac{Q}{2}$
$Q_{2} = Q - \frac{Q}{2} = \frac{Q}{2}$

Q. Total charge Q is broken in two parts Q1​&Q2​ and they are placed at a distance R from each other. The maximum force of repulsion between them will occur when

Q2​=RQ​,Q1​=Q−RQ​

Q2​=4Q​,Q1​=Q−R2Q​

Q2​=4Q​,Q1​=Q−43Q​

Q2​=2Q​,Q1​=Q−2Q​