Question

18.3 Coulomb's Law 10. What is the repulsive force between two pith balls that are 8.00 cm apart and have equal charges of 30.0 nC?

Answer 1

Solution:

Coulomb's law gives the force of attraction or repulsion between two charges which are separated by a distance. It is a vector quantity having both magnitude and direction.

$F = \frac{k q_{1}q_{2}}{r^{2}}$ where q₁ and q₂ are the charges in the bodies

r is the distance between them

k is Coulomb's constant , $k = \frac{1}{4\pi \varepsilon _{0}} = 9\times 10^{9} \frac{Nm^{2}}{C^{2}}$

Here, r = 8.00cm = 8.00 x10^-2 m

q₁ = q₂ = -30 nC = -30 x 10^-9 C

Therefore, $F = \frac{9\times 10^{9}\times 900\times 10^{-18}}{64 \times 10^{-4}}= \frac{8.1\times 10^{-6}}{64\times 10^{-4}} = 1.27\times 10^{-3} N$

Since, both the charges are the same the force is repulsive in nature which is represented by the positive sign