Question

Of the chargeQinitially on a tiny sphere, a portion qis to be transferred to a second, nearby sphere. Both spheres can be treated as particles and are fixed with a certain separation. For what value of q/Qwill the electrostatic force between the two spheres be maximized?

Short answer

Answer:

The value of q/Q for which the electrostatic force between the two spheres will be maximized is 0.5 .

Step-by-step solution

The given data

The charge initially of the first sphere is Q and a portion is q transferred to the nearby sphere.

Understanding the concept of induction

Due to induction, the charged body induces some charge on the nearby uncharged body. Now, using the concept of the Coulomb force of attraction or repulsion, we can get the force applied by the first sphere of the other sphere due to their charges. Now, for the maximum value of the ratio of the charges, we take the differentiation of the force to be zero.

Formula:

The electrostatic force of attraction or repulsion on a body by another charged body according to Coulomb’s law, $F=\frac{k{q}_1{q}_2}{r^2}$ (i)

Calculation of the ratio of the charges for maximum electrostatic force

As per the given data, if q is transferred by the first sphere to the second sphere, then the charge remaining on the first sphere is (Q-q).

Let,r be the separation between the centres of the two spheres.

Then, the electrostatic force between them can be given using equation (i) as follows:

$F=\frac{kq\left(Q-q\right)}{r^2}$

Now, for the case of maximum force, the value of the ratio q/Q can be calculated as follows:

$$\begin{gathered} \frac{dF}{dq}=0 \\ \frac{k}{r^2}\frac{d\left(q\left(Q-q\right)\right)}{dq}=0 \\ \left(Q-q\right)-q=0 \\ Q-2q=0 \end{gathered}$$

Solving further,

$$\begin{gathered} Q=2q \\ \frac{q}{Q}=\frac{1}{2} \\ \frac{q}{Q}=0.5 \end{gathered}$$

Hence, the ratio of q/Q is 0.5.