A charge q is divided into two parts q₁, and (q – q₁) what is the ratio of q/q₁ so that the force between the two parts placed a given distance apart is maximum?

Concept: Given one part of the charge is q₁. Thus other part of the charge is q - q₁.

From Columb's law: 

Electrostatic force b/w them,

\(F = \frac{1}{4\pi \epsilon_0} \ \frac{q_1(q - q_1)}{r^2}\)

or, \(F = K\frac{q_1(q - q_1)}{r^2}\)

where \(K = \frac{1}{4\pi \epsilon_0}\) = 9 × 10⁹ = constant

r = distance between two charges 

For maximum charge

\(\frac{dF}{dq_1}=0\)

∴ \(\frac{d}{dq_1}\left(\frac{kq_1(q-q_1)}{r^2}\right)\) = 0

or, \(\frac{d}{dq_1}(qq_1 - q_1^2)\) = 0

or, q - 2q₁ = 0

or, q = 2q₁

or, q₁ = \(\frac{q}{2}\)

∴ q - q₁ = q - \(\frac{q}{2} = \frac{q}{2}\)

∴ Force b/w them is maximum if the two parts have equal change \(\frac{q}{2}\) each.

∴ The ratio of \(\frac{q}{q_1}\) = 2 ∶ 1

Therefore, current option is 2.