F = 0.06N. Since the charges has different signs the force of atraction between them is 0.06N.
In order to solve this exercise we have to use Coulomb's Law equation which says that the magnitude of each of the electric forces with which two point charges at rest interact is directly proportional to the product of the magnitude of both charges and inversely proportional to the square of the distance that separates them. Given by the equation:
[tex]F = k\frac{|q_{1}q_{2}|}{d^{2}}[/tex]. Where k is the Coulomb's Constant [tex]k = 9x10^{9}\frac{Nm^{2} }{C^{2} }[/tex], q1 and q2 are the charges value in Coulomb (C), and d is the distance between charges in meters (m).
There are two balloons of charges +3.37 x 10-6 C and –8.21 x 10-6 C. The distance between the two balloons is 2.00 m. Calculate the force between the two ballons.
[tex]F = 9x10^{9}\frac{Nm^{2} }{C^{2} } \frac{|(3.37x10^{-6}C)(-8.21x10^{-6}C)|}{(2.00m)^{2}}\\F = 9x10^{9}\frac{Nm^{2} }{C^{2} } \frac{|-2.77x10^{-11}C^{2} )|}{4.00m^{2}}\\F = 9x10^{9}\frac{Nm^{2} }{C^{2} }(6.92x10^{-12})\frac{C^{2} }{m^{2}}\\F = 0.06N[/tex]
