Question

Two pellets, with charges q1 = 22 μC and q2 = 11 μC (1 μC = 10-6 C or 1 microCoulomb = 0.000001 Coulombs), are located 0.52 m apart.

(a) What is the electric force between them? in N(b) What mass object would experience this same force in the Earth's gravitational field? in kg

Answer 1

Answer:

8.0 N

0.82 kg

Explanation:

Given data

q₁ = 22 × 10⁻⁶ c

q₂ = 11 × 10⁻⁶ c

distance (r): 0.52 m

We can find the electrical force between these charges using he Coulomb's law.

$F=k_{e}.\frac{q_{1}.q_{2}}{r^{2}} =8.99\times 10^{9} N.m^{2}.c^{-2} .\frac{22 \times 10^{-6}c . 11 \times 10^{-6}c .}{(0.52m)^{2} } =8.0N$

If the same force were experienced due to the gravitational field, we could find the value of mass associated with it using Newton's second law of motion.

F = m . g

8.0 N = m . 9.8 m/s²

m = 0.82 kg