Question

Two insulated charged copper spheres A and B have their centres separated by a distance of 50 cm. What is the mutual force of electrostatic repulsion if the charge on each is $\sf \: 6.5 \times {10}^{ - 7} C$? The radii of A and B are negligible compared to the distance of separation.

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Answer 1

Answer:

(a) 

Charge on sphere A, qA=6.5×10−7C 

Charge on sphere B, qB=6.5×10−7C

Distance between the spheres, r=50cm=0.5m

Force of repulsion between the two spheres,

F=4π∈0r2qAqB

Where, ∈0= Free space permittivity

4π∈01=9×109Nm2C2

∴F=(0.5)29×109×(6.5×10−7)2

         =1.52×10−2N

Therefore, the force between the two spheres is 1.52×10−2N

(b) 

After doubling the charge, 

Charge on sphere A, qA=2×6.5×10−7C=1.3×10−6C 

Charge on sphere B, qB=2×6.5×10

Answer 2

Answer:

6500

Explanation:

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