If Q = 20 μC and L = 60 cm, what is the magnitude of the electrostatic force on any one of the charges shown?
A) 25 N
B) 19 N
C) 15 N
D) 9.1 N
E) 14 N

Question

If Q = 20 μC and L = 60 cm, what is the magnitude of the electrostatic force on any one of the charges shown? 
A) 25 N 
B) 19 N 
C) 15 N 
D) 9.1 N 
E) 14 N

Quizplus · Accepted Answer

The electrostatic force between two point charges is given by Coulomb's law: 
$$F = \frac{kQq}{r^2}$$ 
where $k$ is Coulomb's constant ($9	imes10^9$ Nm$^2$/C$^2$), $Q$ and $q$ are the magnitudes of the charges, and $r$ is the distance between the charges. 
In this problem, we have $Q=20 \mu$C $= 20 	imes 10^{-6}$ C and $L=60$ cm $= 0.6$ m. 
Let's consider the force acting on the charge in the center. The charges on either side will exert equal and opposite forces on the center charge, so their contributions will cancel out. The force between the center charge and one of the other charges is: 
$$F = \frac{kQ^2}{r^2}$$ 
where $r$ is the distance from the center charge to one of the other charges. 
Since the two charges on either side are equidistant from the center charge, we can treat them as point charges with magnitude $q=Q/2 = 10 \mu$C. The distance $r$ is half the total length, so $r = L/2 = 0.3$ m. 
Plugging in the numbers: 
$$F = \frac{(9	imes10^9)(20	imes10^{-6})^2}{(0.3)^2} \approx 9.1 	ext{ N}$$ 
Therefore, the magnitude of the electrostatic force on any one of the charges is 9.1 N.

If Q = 20 μC and L = 60 Cm