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Physics Question on Electric Charges and Coulomb's Law

(a) 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 6.5×10−7C6.5 × 10^{−7} C? The radii of A and B are negligible compared to the distance of separation.
(b) What is the force of repulsion if each sphere is charged double the above amount, and the distance between them is halved?

Answer

(a) Charge on sphere A, qA=6.5×10−7Cq_A = 6.5 × 10^{−7} C
Charge on sphere B, qB=6.5×10−7Cq_B = 6.5 × 10^{−7 }C
Distance between the spheres,r=50cm=0.5m r = 50 cm = 0.5 m
Force of repulsion between the two spheres
F=14πε0.qAqBr2F = \frac{1}{ 4πε_0}.\frac{q_Aq_B}{ r^2}
Where, ε0ε_0 = Permittivity of free space and 14πε0=9×109Nm2C−2\frac{1}{ 4πε_0} = 9 × 10^9 Nm^2C^{-2}
Therefore,
F=9×109×(6.5×10−7)2(0.5)2F =\frac{ 9 × 10^9 × (6.5 × 10 ^{-7} )^ 2 }{(0.5)^2}
=1.52×10−2N= 1.52 × 10^{-2} N
Therefore, the force between the two spheres is 1.52×10−2N.1.52 × 10^{−2} N.

(b) After doubling the charge,
Charge on sphere A, qA=1.3×10−6Cq_A = 1.3 × 10^{−6} C
Charge on sphere B, qA=1.3×10−6C q_A = 1.3 × 10^{−6} C
The distance between the spheres is halved.
r=0.52=0.25mr = \frac{0.5}{ 2} = 0.25 m
Force of repulsion between the two spheres,
F=14πε0.qAqBr2=9×109×1.3×10−6×1.3×10−6(0.25)2F = \frac{1}{4πε_0}.\frac{q_Aq_B }{ r^2} = \frac{9 × 10^ 9 × 1.3 × 10^{-6}× 1.3× 10^{-6} } {(0.25)^2}
=16×1.52×10−2= 16 × 1.52 × 10^{−2}
=0.243N= 0.243 N
Therefore, the force between the two spheres is 0.243 N.