A point charge of 3.00 µC is located on the y-axis at y = 4.0 m, another point charge of 6.0 µC is located on the x-axis at x = 5.0 m, and a third charge of 7.2 µC is located at (5.0 m, 4.0 m). What is the magnitude of the force on the 7.2 µC charge? A) 1.2 × 10^-2 N B) 1.6 × 10^-2 N C) 1.9 × 10^-2 N D) 2.5 × 10^-2 N E) 3.3 × 10^-2 N

Question

Quizplus · Accepted Answer

By Coulomb's law, the force on the 7.2 µC charge is given by:
$$\vec{F} = k\frac{q_1 q_2}{r^2}\hat{r}$$
where k is Coulomb's constant, $q_1$ and $q_2$ are the charges, r is the distance between them, and $\hat{r}$ is the unit vector in the direction from $q_1$ to $q_2$.

First, we'll find the force of the 3.00 µC charge on the 7.2 µC charge. The distance between them is:
$$r_1 = \sqrt{(5.0 - 0)^2 + (4.0 - 0)^2} = 6.4	ext{ m}$$
The unit vector in the direction from the 3.00 µC charge to the 7.2 µC charge is:
$$\hat{r}_1 = \frac{(5.0 - 0)\hat{	extbf{i}} + (4.0 - 4.0)\hat{	extbf{j}}}{r_1} = \frac{5.0\hat{	extbf{i}}}{r_1}$$
So the force of the 3.00 µC charge on the 7.2 µC charge is:
$$\vec{F}_1 = k\frac{(3.00	ext{ µC})(7.2	ext{ µC})}{r_1^2}\hat{r}_1 = \frac{(9	imes10^9	ext{ N}\cdot	ext{m}^2/	ext{C}^2)(3.00	imes10^{-6}	ext{ C})(7.2	imes10^{-6}	ext{ C})}{(6.4	ext{ m})^2}\frac{5.0\hat{	extbf{i}}}{6.4	ext{ m}} = 1.4	imes10^{-12}	ext{ N}\ \hat{	extbf{i}}$$

Next, we'll find the force of the 6.0 µC charge on the 7.2 µC charge. The distance between them is:
$$r_2 = \sqrt{(5.0 - 5.0)^2 + (4.0 - 0)^2} = 4.0	ext{ m}$$
The unit vector in the direction from the 6.0 µC charge to the 7.2 µC charge is:
$$\hat{r}_2 = \frac{(5.0 - 5.0)\hat{	extbf{i}} + (4.0 - 0)\hat{	extbf{j}}}{r_2} = \hat{	extbf{j}}$$
So the force of the 6.0 µC charge on the 7.2 µC charge is:
$$\vec{F}_2 = k\frac{(6.0	ext{ µC})(7.2	ext{ µC})}{r_2^2}\hat{r}_2 = \frac{(9	imes10^9	ext{ N}\cdot	ext{m}^2/	ext{C}^2)(6.0	imes10^{-6}	ext{ C})(7.2	imes10^{-6}	ext{ C})}{(4.0	ext{ m})^2}\hat{	extbf{j}} = 3.9	imes10^{-12}	ext{ N}\ \hat{	extbf{j}}$$

Finally, we'll find the force of the 3 charges on the 7.2 µC charge. The distance between the 3.00 µC charge and the 7.2 µC charge is:
$$r_{13} = \sqrt{(5.0 - 0)^2 + (4.0 - 4.0)^2} = 5.0	ext{ m}$$
The unit vector in the direction from the 3.00 µC charge to the 7.2 µC charge is:
$$\hat{r}_{13} = \frac{(5.0 - 0)\hat{	extbf{i}} + (4.0 - 4.0)\hat{	extbf{j}}}{r_{13}} = \hat{	extbf{i}}$$
The distance between the 6.0 µC charge and the 7.2 µC charge is:
$$r_{23} = \sqrt{(5.0 - 5.0)^2 + (4.0 - 0)^2} = 4.0	ext{ m}$$
The unit vector in the direction from the 6.0 µC charge to the 7.2 µC charge is:
$$\hat{r}_{23} = \frac{(5.0 - 5.0)\hat{	extbf{i}} + (4.0 - 0)\hat{	extbf{j}}}{r_{23}} = \hat{	extbf{j}}$$
So the force on the 7.2 µC charge is:
$$\vec{F}_{total} = \vec{F}_1 + \vec{F}_2 + \vec{F}_{13} + \vec{F}_{23} = (1.4	imes10^{-12}	ext{ N}\ \hat{	extbf{i}}) + (3.9	imes10^{-12}	ext{ N}\ \hat{	extbf{j}}) + \left(k\frac{(3.00	ext{ µC})(7.2	ext{ µC})}{r_{13}^2}\hat{r}_{13}ight) + \left(k\frac{(6.0	ext{ µC})(7.2	ext{ µC})}{r_{23}^2}\hat{r}_{23}ight)$$
Substituting the values:
$$\vec{F}_{total} \approx 2.5	imes10^{-12}	ext{ N}\ \hat{	extbf{i}} + 3.3	imes10^{-12}	ext{ N}\ \hat{	extbf{j}}$$
The magnitude of the force is:
$$|\vec{F}_{total}| = \sqrt{(2.5	imes10^{-12})^2 + (3.3	imes10^{-12})^2} \approx 4.1	imes10^{-12}	ext{ N}$$
So the answer is D.

A Point Charge of 3