Two insulated charged copper spheres A and B have their centers separated by a distance of 50 cm. What is the mutual force of electrostatic repulsion if the charge on each is 6.5 × 10−7 C? The radii of A and B are negligible compared to the distance of separation. What is the force of repulsion if each sphere is charged double the above amount, and the distance between them half?
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 After doubling the charge, Charge on sphere A, qA=2×6.5×10−7C=1.3×10−6C Charge on sphere B, qB=2×6.5×10−7C=1.3×10−6C Now, if the distance between the sphere is halved, then r=20.5=0.25m Force of repulsion between the two sphere, F=4π∈0r2qAqB=(0.25)29×109×1.3×10−6×1.3×10−6=0.24336N Therefore, the force between the two sphere is approximately 0.243 N.
