Question

calculations:

1. three point - charges are placed on the x - axis. a charge of + 2.0 μc is placed at the origin, - 2.0 μc to the right at x = 50 cm, and + 4.0 μc at the 100 cm mark. what are the magnitude and direction of the electrostatic force which acts on the charge at the origin?

solution

Explanation:

Step1: Recall Coulomb's law

The formula for the electrostatic force between two point - charges is $F = k\frac{q_1q_2}{r^{2}}$, where $k = 9\times10^{9}\ N\cdot m^{2}/C^{2}$, $q_1$ and $q_2$ are the charges, and $r$ is the distance between them.

Step2: Calculate the force due to the charge at $x = 50\ cm$

Let $q_1=2.0\times10^{- 6}\ C$, $q_2=-2.0\times10^{-6}\ C$, and $r_1 = 0.5\ m$. Then $F_1=k\frac{q_1q_2}{r_1^{2}}=(9\times10^{9})\frac{(2.0\times10^{-6})( - 2.0\times10^{-6})}{(0.5)^{2}}=- 0.144\ N$. The negative sign indicates an attractive force.

Step3: Calculate the force due to the charge at $x = 100\ cm$

Let $q_1 = 2.0\times10^{-6}\ C$, $q_3=4.0\times10^{-6}\ C$, and $r_2 = 1.0\ m$. Then $F_2=k\frac{q_1q_3}{r_2^{2}}=(9\times10^{9})\frac{(2.0\times10^{-6})(4.0\times10^{-6})}{(1.0)^{2}}=0.072\ N$.

Step4: Find the net force on the charge at the origin

The net force $F_{net}=F_1 + F_2=-0.144+0.072=- 0.072\ N$.

Answer:

The magnitude of the electrostatic force on the charge at the origin is $0.072\ N$, and the direction is towards the positive $x$ - axis.