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:
the answer for the question is the last option
I am using the equation F=ma (force equals mass times acceleration) to solve these problems.
1. You are looking for force, and have mass and acceleration. You just plug in the values for mass and acceleration to get the force needed.
F=(15kg)(5m/s^2)
F=75N
2. Again, you are looking for force, and just need to plug in the values for mass and acceleration
F=(3kg)(2.4m/s^2)
F=7.2N
3. In this problem, you have force and mass, but need to find acceleration. To do this, you need to get acceleration alone on one side of the equation - divide each side by m. Your equation will now be F/m=a
a=(5N)/(3.7kg)
a=18.5m/s^2
I did not use significant figures. Let me know if you need to do that and need any help on that. Hope this helps!
Do find the percentage you:
5.43/15.6=0.34
Time by 100:
34%
Closest answer is 35%
Hope this helps ;)
