Question

Two 2.00 mC point charges are located on the x-axis. One is at x=1.00 (a) Determine the electric field on the y-axis at y=0.500 m (b) Calculate the electric force on a -3.00C charge placed on the y-axis at y=0.500 m

Answer 1

(a) Let's draw a free-body diagram:

Let the first 2.00 mC point charge is located at a distance $x=1.0\ m$ from the origin and the second 2.00 mC point charge - at a distance $x=-1.0\ m$. Let Q be the point where we need to determine the electric field due to two point charges and let $y=0.5\ m$ be the distance to that point. As we know, the electric field directed away from the positive electric charges. So, we can denote both fields due to our positive charges as $E_+$. Let's draw the $x$- and $y$-components of electric field $E_+$. As we can see from the FBD, $E_{+x}$ components will canceled. So, we have only two $E_{+y}$ components of the electric field.

We can find the magnitude of the electric field from the formula:

$$E_+=k\dfrac{q}{r^2},$$

here, $k=8.99\cdot 10^9\ N\dfrac{m^2}{C^2}$ is the Coulomb's constant, $q=2\cdot 10^{-6} C$ is the charge, $r$ is the distance between the charge and point Q.

We can find the distance between the charge and point Q from the Pythagorean theorem:

$$r=\sqrt{x^2+y^2}=\sqrt{(1.0\ m)^2+(0.5\ m)^2}=1.12\ m.$$

Since we deal only with $E_{+y}$ components we can multiply E_+ by $sin\theta$. We can find angle $\theta$ from the geometry:

$$sin\theta=\dfrac{y}{r}=\dfrac{0.5\ m}{1.12\ m}.$$

Finally, we can find the net electric field on the y-axis at y=0.500 m due to two point charges:

$$E_{net}=2E_{+y}=2E_+sin\theta=2k\dfrac{q}{r^2}\dfrac{y}{r},$$ $$E_{net}=2\cdot 8.99\cdot 10^9\ N\dfrac{m^2}{C^2}\dfrac{2\cdot 10^{-6} C}{(1.12\ m)^2}\dfrac{0.5\ m}{1.12\ m}=1.28\cdot10^4\ \dfrac{N}{C}.$$

(b) We can calculate the magnitude of the electric force from the formula:

$$F=E_{net}q=1.28\cdot10^4\ \dfrac{N}{C}\cdot 3.0\cdot 10^{-6}\ C=3.84\cdot 10^{-2}\ N.$$

Answer:

(a) $E_{net}=1.28\cdot10^4\ \dfrac{N}{C}.$

(b) $F=3.84\cdot 10^{-2}\ N.$