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Question: Two small conducting spheres of equal radius have charges \(+ 10\mu C\) and \(- 20\mu C\) respective...

Two small conducting spheres of equal radius have charges +10μC+ 10\mu C and 20μC- 20\mu C respectively and placed at a distance RR from each other experience forceF1F_{1}. If they are brought in contact and separated to the same distance, they experience forceF2F_{2}. The ratio of F1F_{1} to F2F_{2} is

A

1 : 8

B

– 8 : 1

C

1 : 2

D

– 2 : 1

Answer

– 8 : 1

Explanation

Solution

FQ1Q2F \propto Q_{1}Q_{2}F1F2=Q1Q2Q1Q2=10×205×5=81\frac{F_{1}}{F_{2}} = \frac{Q_{1}Q_{2}}{Q_{1}^{'}Q_{2}^{'}} = \frac{10 \times - 20}{- 5 \times - 5} = - \frac{8}{1}