Two insulated charged copper spheres A and B have their centres separated by a distance of 50 cm. What is mutual force of repulsion if charge on each sphere is 6.5 x 10^–7C. The radii of A and B are negligible compared to the distance of separation.
What is force of repulsion if
(i) each sphere is charged double the above amount and distance between them is halved?
(ii) the two spheres are placed in water of dielectric constant 80?

Solution

Charge on sphere A = charge on sphere B = 6.5 \times 10^{- 7} C distance between the spheres - r = 50 cm = 0.5 m Force of repulsion as per Coulomb' s law is F = k . \frac{q_{1} q_{2}}{r^{2}} = \frac{9 \times 10^{9} \times (6.5)^{2} \times (10^{- 7})^{2}}{(0.5)^{2}} = 1.52 \times 10^{- 2} N a) if each sphere is charged double and the distance between them is halved Charge on sphere A = Charge on sphere B = 2 \times (6.5 \times 10^{- 7}) = 13 \times 10^{- 7} distance between the spheres - r = 0.25 m Force of repulsion - F = k . \frac{q_{1} q_{2}}{r^{2}} = \frac{9 \times 10^{9} \times 4 \times (6.5)^{2} \times (10^{- 7})^{2}}{(0.25)^{2}} = 0.243 N

b) the two spheres are placed in water of dielectric constant 80 i.e, k=80

F = \frac{1}{4 {πε}_{\circ} K} \frac{q_{1} q_{2}}{r^{2}} = \frac{F_{air}}{K} = \frac{1.52 \times 10^{- 2}}{80} = 1.9 \times 10^{- 4} N

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