Question

A charge Q is to be divided on two subjects.
The values of the charges on the objects so
that the force between the objects can be
maximum are
2QQ
(a)
(b)
3 3
4 '4
(d) 2,0
22
units and 2
3Q q
(c​

Answer 1

Given:

A charge Q has to be divided on two two subjects such that the electrostatic force between the objects become maximum.

To find:

Division of the charges ?

Concept:

We shall use differentiation in calculus to find out the division of charge in this case.

Calculation:

Let one charge be q , then other charge will be (Q - q).

Electrostatics force be F and k be Coulomb's Constant .

$F = \dfrac{k(q)(Q - q)}{ {d}^{2} }$

$= > F = \dfrac{k \{Qq - {q}^{2} \} }{ {d}^{2} }$

Differentiation of this function wrt to q :

$= > \dfrac{dF}{dq} = \dfrac{k}{ {d}^{2} } \bigg \{ \dfrac{d(Qq - {q}^{2} )}{dq} \bigg \}$

$= > \dfrac{dF}{dq} = \dfrac{k}{ {d}^{2} } \bigg \{ Q - 2q \bigg \}$

Now , for maxima , dF/dq should be equal to zero .

$\therefore \: \dfrac{dF}{dq} = \dfrac{k}{ {d}^{2} } \bigg \{ Q - 2q \bigg \} = 0$

$= > \dfrac{k}{ {d}^{2} } \bigg \{ Q - 2q \bigg \} = 0$

$= > Q - 2q = 0$

$= > 2q = Q$

$= > q = \dfrac{Q}{2}$

So the charges should be divided equally to get the maximum Electrostatic Force .