Question

Two equal positive point charges q1=q2=2µC interact with a third point charge q3=4µC. Find

the magnitude and direction of the total (net) force on q3. (q1 is 0.3m above the origin; q2 is

0.3m below the origin; q3 is 0.4m right from the origin).

Answer 1

Let's write the forces that act on charge $q_3$ due to charges $q_1$and $q_2$:

$$F_1=\dfrac{kq_1q_3}{r^2},$$ $$F_2=\dfrac{kq_2q_3}{r^2}.$$

We can find the distance from the charges $q_1$and $q_3$ as well as $q_2$ and $q_3$ from the Pythagorean theorem:

$$r=\sqrt{(0.3\ m)^2+(0.4\ m)^2}=0.5\ m.$$

Since, $q_1=q_2$ and $r_1=r_2=r$ the magnitudes of two forces are equal. Let's write the components of forces $F_1$ and $F_2$ in projections on $x$ and $y$ axis:

$$F_1cos\theta+F_2cos\theta=2F_1cos\theta,$$ $$-F_1sin\theta+F_2sin\theta=0.$$

As we can see, $y$-components cancel out, so we can write the net force on $q_3$:

$$F_{net}=2\dfrac{kq_1q_3}{r^2}cos\theta.$$

We can find angle $\theta$ from the geometry:

$$\theta=cos^{-1}(\dfrac{0.4\ m}{0.5\ m})=36.86^{\circ}.$$

Finally, we can find the net force on $q_3$:

$$F_{net}=2\cdot\dfrac{9\cdot10^9\ \dfrac{Nm^2}{C^2}\cdot(2\cdot10^{-6}\ C)^2}{(0.5\ m)^2}cos36.86^{\circ}=0.23\ N.$$

Answer:

$F_{net}=0.23\ N,$ in $+x$-direction.