Repulsive force of magnitude 6 × 10−3 N Charge on the first sphere, q1 = 2 × 10−7 C Charge on the second sphere, q2 = 3 × 10−7 C econd Distance between the spheres, r = 30 cm = 0.3 m Electrostatic force between the spheres is given by the relation,

Where, ∈0 = Permittivity of free space

Hence, force between the two small charged spheres is 6 × 10−3 N. The charges are of same nature. Hence, force between them will be repulsive.

Question 1.2: The electrostatic force on a small sphere of charge 0.4 μC due to another small sphere of between 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? Answer

Electrostatic force on the first sphere, F = 0.2 N Charge on this sphere, q1 = 0.4 μC = 0.4 × 10−6 C Charge on the second sphere, q2 = − 0.8 μC = − 0.8 × 10−6 C Electrostatic force between the spheres is given by the relation,

Where, ∈0 = Permittivity of free space

The distance between the two spheres is 0.12 m. 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.

Question 1.3: Check that the ratio ke2/G memp is dimensionless. Look up a Table of Physical Constants and determine the value of this ratio. What does the ratio signify? Answer

The given ratio is Where, G = Gravitational constant Its unit is N m2 kg−2.

.

me and mp = Masses of electron and proton. Their unit is kg. e = Electric charge. Its unit is C.

Its unit is N m2 C−2.

∈0 = Permittivity of free space

Hence, the given ratio is dimensionless. e = 1.6 × 10−19 C G = 6.67 × 10−11 N m2 kg-2 me= 9.1 × 10−31 kg mp = 1.66 × 10−27 kg Hence, the numerical value of the given ratio is

This is the ratio of electric force to the gravitational force between a proton and an electron, keeping distance between