Question

Two small conducting sphere of equal radius have charges \(+1 \mathrm{c}\) and \(-2 \mathrm{c}\) respectively and placed at a distance \(\mathrm{d}\) from each other experience force \(F_{1}\). If they are brought in contact and separated to the same distance, they experience force \(F_{2}\). The ratio of \(F_{1}\) to \(F_{2}\) is \(\ldots \ldots \ldots \ldots\) (A) \(-8: 1\) (B) \(1: 2\) (C) \(1: 8\) (D) \(-2: 1\)

Short answer

The ratio of \(F_1\) to \(F_2\) is \(-8:1\).

Step-by-step solution

Calculate force F1

Use Coulomb's Law formula for the force between two charged objects: \(F_1 = k \frac{Q_1 Q_2}{d^2}\) where, k is Coulomb's constant = \(9 \times 10^9 Nm^2C^{-2}\), \(Q_1 = +1c\), \(Q_2 = -2c\), and d is the distance between the spheres. Plug in the values: \(F_1 = 9 \times 10^9 \frac{(+1)(-2)}{d^2}\)

Determine redistributed charges

When the spheres come into contact, the total charge becomes, \(Q_{total} = Q_1 + Q_2 = +1c - 2c = -1c\) Since the spheres have equal radius, the charge will be distributed equally between them after touching, so each sphere gets half of the total charge: \(Q_{1(new)} = Q_{2(new)} = -\frac{1}{2}c\)

Calculate force F2

Use Coulomb's Law formula again, now with the redistributed charges of the spheres. \(F_2 = k \frac{Q_{1(new)} Q_{2(new)}}{d^2}\) Plug in the values: \(F_2 = 9 \times 10^9 \frac{(-\frac{1}{2})(-\frac{1}{2})}{d^2}\)

Calculate the ratio of F1 to F2

Now, we need the ratio of F1 to F2. We can cancel the common factors on both sides, so we get: \(\frac{F_1}{F_2} = \frac{[-2]}{[\frac{1}{4}]}\) Solve it: \(\frac{F_1}{F_2} = -8\)

Select the correct option

The ratio of F1 to F2 is -8:1. So, the correct answer is: (A) -8:1