Q. 24.3( 134 Votes )

# 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?

Answer :

(a) The electrostatic force, F = 0.2 N

Charge on this present sphere, q_{1} = 0.4 μC = 0.4 × 10^{-6} C

Charge on the other sphere, q_{2} = –0.8 μC = -0.8 × 10^{-6} C

The electrostatic force between the spheres can be given by the equation,

…(1)

Where, q_{1} and q_{2} are the charges.

r is the distance between the charges.

is a constant and its value is 9x10^{9} N m^{2} C^{-2}

ϵ_{0} is the permittivity of free space. Its value is 8.85 x 10^{-12} (F/m)

Therefore, from equation (1),

⇒ r^{2} = 144 x 10^{-4} m^{2}

Taking square root both the sides,

⇒ r =

r = 0.12 m

Hence, the distance between the two spheres is 0.12 m.

(b) Since the nature of the charges on the spheres is opposite, the force between them will be attractive in nature. The spheres will attract each other with the same amount of force.

Therefore, the force on the second sphere due to the first sphere will be 0.2 N.

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