Question

A charge q is divided into two parts q and $\left( {Q - q} \right).$ What is q in terms of $Q$ when repulsion between them is maximum
A. $\dfrac{Q}{2}$
B. $Q$
C. $2Q$
D. $\dfrac{2}{Q}$

Answer 1

- Should know concepts of Coulomb force of attraction and repulsion.
- Should know differentiation.

Complete step by step Solution:
Coulomb's law states that the electrical force between two charged particles is directly proportional to the product of the quantity of charge on the objects and inversely proportional to the square of
the separation distance between the two charges.

$F = \dfrac{{k{Q_1}{Q_2}}}{{{d^2}}}$

Where ${\mathbf{Q}}$ 1 represents the quantity of charge on object 1$\left( {in{\text{ }}Coulombs} \right),\;{{\mathbf{Q}}_{\mathbf{2}}}$ represents the quantity of charge on object 2 $\left( {in{\text{ }}Coulombs} \right),$ and d represents the distance of separation between the two objects $\left( {in{\text{ }}meters} \right).$The symbol k is a proportionality constant known as the Coulomb's law constant. The value of this constant is dependent upon the medium that the charged objects are immersed in. In the case of air, the value is approximately,

$K =$ $9.0{\text{ }} \times {10^9}\;\dfrac{{N{\text{ }}{m^{2\;}}}}{{{C^2}}}$.

For a given distance between two charges the attraction$/$ repulsion will be maximum, when according to Coulomb's law , the product of charges is maximum.

In the present problem charges are $q{\text{ }}and{\text{ }}\left( {Q - q} \right).$ The product is $q\left( {Q - q} \right).$

Let parts are separated at a distance of

So repulsion force $(electrostatic{\text{ }}force{\text{ }}of{\text{ }}attraction),$

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

Here, Q and d are constant so for maximum force differentiate the force with respect to q,

$$\dfrac{{dF}}{{dq}} = \dfrac{d}{{dq}}\left[ {\dfrac{{kq\left( {Q - q} \right)}}{{{d^2}}}} \right] \\\ \Rightarrow \dfrac{{dF}}{{dq}} = \dfrac{k}{{{d^2}}}\dfrac{d}{{dq}}\left[ {q\left( {Q - q} \right)} \right] \\\ \Rightarrow \dfrac{{dF}}{{dq}} = \dfrac{k}{{{d^2}}}\dfrac{d}{{dq}}\left[ {qQ - {q^2}} \right] \\\ \Rightarrow \dfrac{{dF}}{{dq}} = \dfrac{k}{{{d^2}}}\left( {Q - 2q} \right) \\\$$

For maximum force,
$\dfrac{{dF}}{{dq}} = 0$

$\Rightarrow \dfrac{k}{{{d^2}}}\left( {Q - 2q} \right) = 0$

$$\left( {Q - 2q} \right) = 0 \\\ \Rightarrow Q = 2q \\\ \Rightarrow q = \dfrac{Q}{2} \\\$$

So, the correct Option is (A)

Note:
- q is a variable in the given conditions, For any function F to be maximum or minimum the differential of the function with respect to the given variable $\left( {here{\text{ }}it{\text{ }}is{\text{ }}q} \right)$ is equal to zero.
- When the differential is equated to zero, you will get the value of $q$ at which function F is found to be maximum.