Question

Two identical small spheres carry charge of ${Q_1}\& {Q_2}$ with ${Q_1} > > {Q_2}$ . The charges are d distance apart. The force they exert on one another is ${F_1}$ . The spheres are made to touch one another and then separated to distance d apart. The force they exert on one another now is ${F_2}$. Then ${F_1}/{F_2}$ is
A.$$$$4{Q_1}/{Q_2}
$B.\,\dfrac{{{Q_1}}}{{4{Q_2}}}$
$C.\,\dfrac{{4{Q_2}}}{{{Q_1}}}$
$D.\,\dfrac{{{Q_2}}}{{4{Q_1}}}$

Answer 1

Here we have to know the Coulomb’s Law to solve this question. This law provides the force of attraction or repulsion between two point charges. If there is ${Q_1}$and ${Q_2}$ two point charges and they are separated by the distance r. then the magnitude of the force of attraction or repulsion between them will be $F = K\dfrac{{\left| {{Q_1}} \right|\left| {{Q_2}} \right|}}{{{r^2}}}$ .where k is constant.

Complete step by step answer:
In the first case according to Coulomb’s Law we know,
${F_1} = K\dfrac{{\left| {{Q_1}} \right|\left| {{Q_2}} \right|}}{{{d^2}}}$
Now, for the second case,
${F_2} = K\dfrac{{\dfrac{{{{({Q_1} + {Q_2})}^2}}}{{{2^2}}}}}{{{d^2}}}$
Here according to the question we know that ${Q_1} > > {Q_2}$
So,

\Rightarrow{F_2} = K\dfrac{{{Q_1}^2}}{{4{d^2}}}$$ Now, $$\dfrac{{{F_1}}}{{{F_2}}} = K\dfrac{{{Q_1}{Q_2}}}{{{d^2}}} \times \dfrac{{4{d^2}}}{{K{Q_1}^2}} \\\ \therefore\dfrac{{{F_1}}}{{{F_2}}}= \dfrac{{4{Q_2}}}{{{Q_1}}}$$ **So the right option would be option number $$A$$.** **Note:** For a continuous charge distribution students need to divide up the charge distribution into infinitesimal pieces and add up the individual forces with the integrals to get net force. In this formula K is called Boltzmann’s constant. The value of this constant is $$8.9876 \times {10^9}\dfrac{{N{m^2}}}{{{C^2}}}$$. We all have to keep it in our mind that if the both charges are same in nature, like if both of them are either positive or negative then the force will be repulsive. And if the charges are opposite in nature as one is positive and other one is negative then the force will be an attractive force.