Question

Two small plastic spheres are given positive electrical charges. When they are 30.0 cm apart, the repulsive force between them has magnitude 0.130 N. If one sphere has four times the charge of the other, what is the charge of the least charged sphere? Give the answer in nanocoulomb (nC).

Answer 1

Charge on least sphere, q = 570 nC

Step-by-step explanation:

It is given that,

Two small plastic spheres are given positive electrical charges. The distance between the spheres, r = 30 cm = 0.3 m

The repulsive force acting on the spheres, F = 0.13 N

If one sphere has four times the charge of the other.

Let charge on other sphere is, q₁ = q. So, the charge on first sphere is, q₂ = 4 q. The electrostatic force is given by :

`F=k(q_1q_2)/(r^2)`

`0.13=9* 10^9* (q* 4q)/((0.3\ m)^2)`

`q^2=(0.13* (0.3)^2)/(9* 10^9* 4)`

`q=5.7* 10^(-7)\ C`

q = 570 nC

So, the charge on the least sphere is 570 nC. Hence, this is the required solution.