Question

Two identical balls having like charges and placed at a certain distance apart repel each other with a certain force. They are brought in contact and then moved apart to a distance equal to half their initial separation. The force of repulsion between them increases $4.5$ times in comparison with the initial value. The ratio of the initial charges of the balls is
A. $4:1$
B. $6:1$
C. $3:1$
D. $2:1$

Answer 1

Hint: A ratio is the quantitative relation between the two amounts that represent the number of times one value is contained in the other. Here, we have taken two balls that are separated by a certain distance. But when they will come in contact with each other, the charge will spread all over the balls.

Complete step by step answer:
Consider two identical balls A and B having like charges that are separated by a certain distance. Therefore, charge on the ball A $ = {Q_1}$.And the charge on the ball B $ = {Q_2}$.
Also, the distance between the ball A and B $ = r$.The balls are placed at a certain distance and they repel each other with a certain force. This force of repulsion acting on the balls is given by
$F = \dfrac{{K{Q_1}{Q_2}}}{{{r^2}}}$
Now, when the balls are brought in contact and then moved apart to a certain distance that will be equal to half of their initial separation. After contact of balls, the amount of charge on the balls will be equal.
Therefore, the magnitude of charge on each ball after contact $ = \dfrac{{{Q_1} + {Q_2}}}{2}$
And the distance between them $ = \dfrac{r}{2}$
Now, the force between the balls after contact, is given by
$F' = \dfrac{{K\left( {{Q_1} + {Q_2}} \right)\left( {{Q_1} + {Q_2}} \right)}}{{4{{\left( {\dfrac{r}{2}} \right)}^2}}}$
$ \Rightarrow \,F' = \dfrac{{K{{\left( {{Q_1} + {Q_2}} \right)}^2}}}{{{r^2}}}$
According to the question,
$ \Rightarrow \,F' = 4.5F$
$ \Rightarrow 4.5 \times \,\dfrac{{K{Q_1}{Q_2}}}{{{r^2}}} = \dfrac{{K{{\left( {{Q_1} + {Q_2}} \right)}^2}}}{{{r^2}}}$
$ \Rightarrow \,9{Q_1}{Q_2} = 2{\left( {{Q_1} + {Q_2}} \right)^2}$
$ \Rightarrow \,9{Q_1}{Q_2} = 2\left( {Q_1^2 + Q_2^2 + 2{Q_1}{Q_2}} \right)$
$ \Rightarrow \,9{Q_1}{Q_2} = 2Q_1^2 + 2Q_2^2 + 4{Q_1}{Q_2}$
$ \Rightarrow \,5{Q_1}{Q_2} = 2Q_1^2 + 2Q_2^2$
Now, putting $\dfrac{{{Q_1}}}{{{Q_2}}} = K$ , we get
$ \Rightarrow \,5 = 2K + \dfrac{2}{K}$
$ \Rightarrow \,5K = 2{K^2} + 2$
$ \Rightarrow \,2{K^2} - 5K + 2 = 0$
Now, factoring the above equation, we get
$2K\left( {K - 2} \right) - 1\left( {K - 2} \right) = 0$
$ \Rightarrow \,\left( {2K - 1} \right)\left( {K - 2} \right) = 0$
$ \Rightarrow \,K = 2,\dfrac{1}{2}$
Now, neglecting $K = \dfrac{1}{2}$ , we get
$K = 2$
$ \Rightarrow \,\dfrac{{{Q_1}}}{{{Q_2}}} = 2$
Therefore, the ratio of the initial changes of the ball is $2:1$ .

Hence, option D is the correct option.

Note:In the above solution, we have taken the charge on each ball after they come in contact as $\dfrac{{{Q_1} + {Q_2}}}{2}$. This is because of the process of conduction. Conduction says that when the two balls come in contact with each other, the charge will spread all over the balls.