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Question: Two charges of \(40\mu C\) and\(- 20\mu C\) are placed at a certain distance apart. They are touched...

Two charges of 40μC40\mu C and−20μC- 20\mu C are placed at a certain distance apart. They are touched and kept at the same distance. The ratio of the initial to the final force between them is

A

8:18:1

B

4:14:1

C

1 : 8

D

1 : 1

Answer

8:18:1

Explanation

Solution

Since only magnitude of charges are changes that’s why F∝q1q2F \propto q_{1}q_{2} ⇒ F1F2=q1q2q1′q2′=40×2010×10=81\frac{F_{1}}{F_{2}} = \frac{q_{1}q_{2}}{q'_{1}q'_{2}} = \frac{40 \times 20}{10 \times 10} = \frac{8}{1}