First slide

Electrostatic force

Question

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

Moderate

A

1:2
B

-8: 1
C

1: 8
D

-2: 1

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

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