Two small conducting sphere of equal radii have charges + 10 $μ$ C and - 20 $μ$ C respectively and placed at a distance R from each other experience force F₁. If they are brought in contact and separated to the same distance, they experience force F₂. The ratio of F₁ to F₂ is

see full answer

Start JEE / NEET / Foundation preparation at rupees 99/day !!

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!

An Intiative by Sri Chaitanya

1:2

-8: 1

1: 8

-2: 1

answer is B.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

Here, $F_{1} = \frac{k (+ 10) (- 20)}{R^{2}} = \frac{- k \times 200}{R^{2}}$

As spheres are of equal radii, so on touching, the net charge = $(+ 10 - 20) μC = - 10 μC$ , is shared equally between them, i.e. each sphere carries $- 5 μC$ charge.

$\begin{array}{l} F_{2} = \frac{k (- 5) (- 5)}{R^{2}} = \frac{k \times 25}{R^{2}} \\ ∵ \frac{F_{1}}{F_{2}} = \frac{- 8}{1} \Rightarrow F_{1} : F_{2} = - 8 : 1 \end{array}$