Question

Two balloons having the same charge of 1.2 × 10-6 coulombs each are kept 5.0 × 10-1 meters apart. What is the magnitude of the repulsive force between them? (k = 9.0 × 109 newton·meter2/coulombs2)
1.2 × 10-2 newtons
3.2 × 10-3 newtons
6.5 × 10-3 newtons
3.0 × 10-4 newtons
5.2 × 10-2 newtons

Answer 1

The answer is 5.2 × 10-2 newtons.

To calculate this, we will use Coulomb's Law:
F = $\frac{k * Q_{1} * Q_{2} }{ r^{2} }$
where F is repulcive force, k is constant, Q is charge, r is distance between charges.

If:
$Q_{1}$ = $Q_{2}$ = 1.2 × $10^{-6}$ C²
r = 5.0 × $10^{-1}$ C²
k = 5.0 × $10^{9}$ $\frac{N*m^{2} }{ C^{2} }$

Then:
$F = \frac{5.0* 10^{-1} \frac{N* m^{2} }{ C^{2}} * 1.2 * 10^{-6} C * 1.2 * 10^{-6} C }{(5.0* 10^{-1} m)^{2}}$
⇒ $F= \frac{9 * 1.2 * 1.2 * 10^{-3} N }{25* 10^{1} }$
⇒$F = \frac{12.96* 10^{-2} }{25} N$
⇒ $F = 5.2 N$

Answer 2

Answer:

For plato users: Option E. 5.2 x 10^-2 is correct.