Class-12th Physics, CBSE and UP Board
Electric Charges and Fields,
Chapter-1, Exercise -1, Q-1.2
NCERT Solutions for Class 12th Physics
The electrostatic force on a small sphere of charge 0.4 μC due to another small sphere of charge –0.8 μC in air is 0.2 N. (a) what is the distance between the two spheres? (b) What is the force on the second sphere due to the first?
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Given the electrostatic force on a small sphere of charge 0.4 µC due to another small sphere of charge – 0.8 µC in air is 0.2 N.
(a) What is the distance between the two spheres?
(b) What is the force on the second sphere due to the first?
(a) Electrostatic force on the first sphere, F = 0.2 N Charge on this sphere, q1 = 0.4 µC = 0.4 X 10-6 C
Charge on the second sphere,
q2 = – 0.8 µC = – 0.8 x 10-6 C
between the spheres is given by the relation
Since Electrostatic force between two charges q1 and q2 separated by a distance of r is given by expression:
F=1 /4πε0 x q1q2/r²
Where ε0 =Permittivity of free space and value of
1 /4πε0 =9 X 109 Nm 2C⁻²
Therefore,
r2= 1 /4πε0 x (q1q2)/F
r2= 9 x 109 x 0.4 x 10-6 x 8 x 10-6 /0.2
r2= 144 x 10-4
r=√(144 x 10-4)=(12×10-2)=0.12
The distance between the two spheres is 0.12 m.
(b) Both the spheres attract each other with the same force. Therefore, the force on the second sphere due to the first is 0.2 N.