Question

Two positive point charges, each of which has a charge of 2.5 × 10−9 C, are located at y = + 0.50m and y = − 0.50m. Find the magnitude and direction of the resultant electric force acting on a charge of 3.0 × 10−9 C located at x = 0.70m

Answer 1

Answer:

The resultant electric force is 14.8N to the right.

Explanation:

Since the three charges aren't in the same line, we have to break down the force in components. First, we need to know the distance from the third charge to the other ones. That is made using the Pythagorean Theorem. As the figure is symmetric with respect to the x-axis, the two distances are the same:

$r=\sqrt{(0.50m)^{2}+(0.70m)^{2}}=0.86m$

Now, we use the Coulomb's Law to obtain the magnitude of the individual forces caused by each charge on the third charge:

$|F_{13}|=k\frac{q_1q_3}{r^{2}} \\\\|F_{13}|=(9*10^{9}Nm^{2}/C^{2})\frac{(2.5*10^{-9}C)(3.0*10^{-9}C)}{(0.86m)^{2}}\\\\|F_{13}|=9.1N$

For the same reason the distances are the same, the magnitude of the forces are the same:

$|F_{23}|=|F_{13}|=9.1N$

So, to get the resultant force, we have to break down this forces in components. To do this, we need their angles with respect to the x-axis. Let θ₁ and θ₂ be these angles, respectively. Then, we calculate them using trigonometry:

$\theta_1=\arctan(\frac{-0.50m}{0.70m})=-35.5\°\\\\\theta_2=\arctan(\frac{0.50m}{0.70m})=35.5\°$

Now, we calculate the components of the forces:

$F_{13}_x=F_{13}\cos\theta_1=9.1N\cos(-35.5\°)=7.4N\\\\F_{13}_y=F_{13}\sin\theta_1=9.1N\sin(-35.5\°)=-5.3N\\\\F_{23}_x=F_{23}\cos\theta_2=9.1N\cos(35.5\°)=7.4N\\\\F_{23}_y=F_{23}\sin\theta_2=9.1N\sin(35.5\°)=5.3N$

Evidently, the y-components cancel out, and the resultant electric force on the third charge is $7.4N+7.4N=14.8N$ along the x-axis (to the right, because it's positive).