Question

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

Answer 1

Answer:

6 × 10 - 3 N.

Explanation:

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

Distance between the spheres, r = 30 cm = 0.3 m

Electrostatic force between the spheres is given by the relation,

F=(q1.q2)/(4pie*∈0 *r^2)

    =6*10^-3 N

Where, ∈0 = Permittivity of free space

Answer 2

$\huge\red{\underline{{\boxed{\textbf{QUESTION}}}}}$

What is the force between two small charged spheres having charges of 2*10^-7 and 3*10^-7 placed 30cm apart in air?

$\huge\red{\underline{{\boxed{\textbf{ANSWER}}}}}$

According to Coulomb's law,

Force between two charged particles in air is given by:

$force = \frac {k(q1 \times q2)}{r}$

$k = \frac{1}{4\pi \: \: \: \: \: \: epsilon}$

= 9 x 10^9 N m² /C²

q1 and q2 = charges of two charged spheres

r = distance between charged spheres

Force between between two small charged spheres charges kept apart in air =

(9 x 10^9 N m² /C²) * (2 x 10^(-7) C) * (3 x 10^(-7) C)/(30 x 10^-2 m)²

= 3/500

= 6 x 10^-3 N

or 6 mN.

Nature of force = Repulsive.

So Force => 6 mN Repulsive.

@HarshPratapSingh