Question

2. What is the charge, expressed in micro coulombs on two equally and similarly charges spheres placed in air with their centres 30 cm apart repelling each other with a force of 120 mgm.

Answer 1

Answer:

q = 0.1086 micro Coulombs

Explanation:

By Coulombs law, we have;

$F =k \times \dfrac{ q_1 \times q_2}{r^2}$

Where;

F = The electric force = 120 mgm

q₁ and q₂ = Charge

r = The separating distance = 30 cm = 0.3 m

k = 8.9876×10⁹ kg·m³/(s²·C²)

Where, q₁ and q₂, we have;

$F =k \times \dfrac{ q^2}{r^2}$

Whereby the force is the force of 120 milligram mass, we have;

0.00012 × 9.81 = 000011772 N

$q = \sqrt{ \dfrac{ F\times r^2}{k}}$

Substituting the values, we have;

$q = \sqrt{ \dfrac{ 000011772 \times (0.3)^2}{8.9876 \times 10^9}} = 1.086 \times 10 ^{-7} \ Coulombs$

q = 0.1086 micro Coulombs.