point charges 1mc and -2mc are located at (3,2,-1) and (-1,-1,4) respectively. calculate the electric force on a -10nc charge located at (0,3,1) and the electric field intensity at that point.

To find the electric force on a -10 nC charge located at (0, 3, 1) due to the point charges of 1 µC and -2 µC, we first need to calculate the electric field produced by each charge at that point. Then, we can determine the force acting on the -10 nC charge using the relationship between electric field and force.

Step 1: Calculate the Electric Field from Each Charge

The electric field \( \mathbf{E} \) created by a point charge \( Q \) at a distance \( r \) is given by the formula:

\( \mathbf{E} = \frac{k \cdot |Q|}{r^2} \cdot \hat{r} \)

where:

• \( k \) is Coulomb's constant, approximately \( 8.99 \times 10^9 \, \text{N m}^2/\text{C}^2 \)
• \( |Q| \) is the magnitude of the charge
• \( r \) is the distance from the charge to the point where the field is being calculated
• \( \hat{r} \) is the unit vector pointing from the charge to the point of interest

Electric Field from the 1 µC Charge

The first charge is 1 µC located at (3, 2, -1). The position vector from this charge to the point (0, 3, 1) is:

\( \mathbf{r}_{1} = (0 - 3, 3 - 2, 1 - (-1)) = (-3, 1, 2) \)

The distance \( r_1 \) is calculated as:

\( r_1 = \sqrt{(-3)^2 + 1^2 + 2^2} = \sqrt{9 + 1 + 4} = \sqrt{14} \)

The unit vector \( \hat{r}_{1} \) is:

\( \hat{r}_{1} = \frac{\mathbf{r}_{1}}{r_1} = \left(-\frac{3}{\sqrt{14}}, \frac{1}{\sqrt{14}}, \frac{2}{\sqrt{14}}\right) \)

Now, substituting into the electric field formula:

\( \mathbf{E}_{1} = \frac{(8.99 \times 10^9) \cdot (1 \times 10^{-6})}{14} \cdot \hat{r}_{1} \)

Calculating \( \mathbf{E}_{1} \):

\( \mathbf{E}_{1} = \frac{8.99 \times 10^3}{14} \cdot \left(-\frac{3}{\sqrt{14}}, \frac{1}{\sqrt{14}}, \frac{2}{\sqrt{14}}\right) \)

After calculating, we find:

\( \mathbf{E}_{1} \approx (-1.93 \times 10^3, 0.64 \times 10^3, 1.29 \times 10^3) \, \text{N/C} \)

Electric Field from the -2 µC Charge

The second charge is -2 µC located at (-1, -1, 4). The position vector from this charge to the point (0, 3, 1) is:

\( \mathbf{r}_{2} = (0 - (-1), 3 - (-1), 1 - 4) = (1, 4, -3) \)

The distance \( r_2 \) is:

\( r_2 = \sqrt{1^2 + 4^2 + (-3)^2} = \sqrt{1 + 16 + 9} = \sqrt{26} \)

The unit vector \( \hat{r}_{2} \) is:

\( \hat{r}_{2} = \frac{\mathbf{r}_{2}}{r_2} = \left(\frac{1}{\sqrt{26}}, \frac{4}{\sqrt{26}}, -\frac{3}{\sqrt{26}}\right) \)

Now substituting into the electric field formula:

\( \mathbf{E}_{2} = \frac{(8.99 \times 10^9) \cdot (2 \times 10^{-6})}{26} \cdot \hat{r}_{2} \)

Calculating \( \mathbf{E}_{2} \):

\( \mathbf{E}_{2} = \frac{17.98 \times 10^3}{26} \cdot \left(\frac{1}{\sqrt{26}}, \frac{4}{\sqrt{26}}, -\frac{3}{\sqrt{26}}\right) \)

After calculating, we find:

\( \mathbf{E}_{2} \approx (0.69 \times 10^3, 2.76 \times 10^3, -2.07 \times 10^3) \, \text{N/C} \)

Step 2: Total Electric Field at the Point

The total electric field \( \mathbf{E}_{\text{total}} \) at the point (0, 3, 1) is the vector sum of \( \mathbf{E}_{1} \) and \( \mathbf{E}_{2} \):

\( \mathbf{E}_{\text{total}} = \mathbf{E}_{1} + \mathbf{E}_{2} \)

Calculating this gives:

\( \mathbf{E}_{\text{total}} \approx (-1.93 \times 10^3 + 0.69 \times 10^3, 0.64 \times 10^3 + 2.76 \times 10^3, 1.29 \times 10^3 - 2.07 \times 10^3) \)

Thus, we find:

\( \mathbf{E}_{\text{total}} \approx (-1.24 \times 10^3, 3.40 \times 10^3, -0.78 \times 10^3) \, \text