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

29.4 uN

Step-by-step 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(q_(1) q_(2) )/(r^(2))`

where k is a proportionality constant known as the Coulomb's law constant. Its value is
`9 * 10^(9)` Nm²/C²

r = distance between charges = 70 cm = 0.7 m

q1 = q2 = 4nC =
`4 * 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 * 10^(9) * (4 * 10^(-9) * 4 * 10^(-9))/(0.7^(2))\\\\ F=2.94*10^(-7) N\\\\ F=29.4 * 10^(-6) N\\\\ F=29.4 \mu N`

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