Question

Two balloons are charged with an identical quantity and type of charge: -4 nc They are held apart at a separation distance of 70 cm. Determine the magnitude of the electrical force of repulsion between them.

Answer 1

Answer:

29.4 uN

Explanation:

The electric force between two charges can be calculated using Coulomb's Law. According to this law the force between two point charges is given as:

$F=k\frac{q_{1} q_{2} }{r^{2}}$

where k is a proportionality constant known as the Coulomb's law constant. Its value is $9 \times 10^{9}$ Nm²/C²

r = distance between charges = 70 cm = 0.7 m

q1 = q2 = 4nC = $4 \times 10^{-9}$ C

The negative sign indicates that the charges are negative. In the formula we will only use the magnitude of the charges.

Using these values in the formula, we get:

$F=9 \times 10^{9} \times \frac{4 \times 10^{-9} \times 4 \times 10^{-9}}{0.7^{2}}\\\\ F=2.94\times10^{-7} N\\\\ F=29.4 \times 10^{-6} N\\\\ F=29.4 \mu N$

Therefore, the magnitude of repulsive force between the given charges will be 29.4 uN