(100, 108)
Due to
1.2x90=108
100, 108
Complete question:
if two point charges are separated by 1.5 cm and have charge values of +2.0 and -4.0 μC, respectively, what is the value of the mutual force between them.
Answer:
The mutual force between the two point charges is 319.64 N
Explanation:
Given;
distance between the two point charges, r = 1.5 cm = 1.5 x 10⁻² m
value of the charges, q₁ and q₂ = 2 μC and - μ4 C
Apply Coulomb's law;
where;
F is the force of attraction between the two charges
|q₁| and |q₂| are the magnitude of the two charges
r is the distance between the two charges
k is Coulomb's constant = 8.99 x 10⁹ Nm²/C²
Therefore, the mutual force between the two point charges is 319.64 N
Answer:
Explanation:
As we know that system of two boxes are moving on frictionless surface
So here if two boxes are considered as a system
then we have
Now since we know that both the boxes are moving together so force applied by first box on other box is given as