Electric Charges & Fields

Q1. What is the force between two small charged spheres having charges 2 × 10-7 C  and 3 × 10-7 C placed 30cm apart in air ?

Sol. Given q1 = 2 × 10-7 C , q2 = 3 × 10-7 C ,

r = 30 cm = 0.3 m , F = ?

By formula , F = 1/4πεo(q1q2 /r2)

= ( 9×109 × 2 × 10-7 × 3 × 10-7 ) /(0.3)2

= 6 × 10-3 N (repulsive)

Q2. 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 ?

Sol. Given , q1 = 0.4 μC  , q2 = −0.8 μC

F12 = 0.2 N

(a) We know that

F = 1/4πεo(q1q2 /r2)

r2 = 1/4πεo(q1q2 /F)

= ( 9×109 × 0.4 × 10-6 × 0.8 × 10-6 ) /(0.2)

= 144 × 10-4

and r = 12 × 10-2 m

= 12 cm

Q3. Check that the ratio ke2/Gmemp is dimensionless. Look up a Table of Physical Constants and determine the value of this ratio. What does the ratio signify ?

Sol. An electron and proton have a charge of 1.6 × 10-19 C each and their masses are 9.1×10-31 kg and 1.67 ×10-27 kg .

Electrostatic force b/w electron and proton ,

Fe = 1/4πεo(e.e/r2)

Gravitational force b/w electron and proton ,

Fg = Gmemp /r2

On dividing ,

Fe/Fg = e2/4πεoGmemp

≈ 2.4 × 1039

The ratio is quite large . This shows that electrostatic forces are much stronger than gravitational forces .

A common example is the lifting of a paper by charged comb against the force of entire Earth on that paper .

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