Understanding Electric Field from Point Charges

This problem involves calculating the total electric field at a specific point in space, generated by two distinct point charges located on the x-axis. We need to find the electric field vector at point P(1, 3) m due to charges $q_1$ and $q_2$. The principle of superposition is key here: the total electric field at P is the vector sum of the electric fields produced by each charge individually.

Given Information and Constants

Let's list the essential values provided in the question:

• Charge $q_1 = \sqrt{10} \, \mu C = \sqrt{10} \times 10^{-6} \, C$. Its position is $x_1 = 0$ m, so its position vector is $\vec{r}_{q1} = (0, 0)$ m.
• Charge $q_2 = -18\sqrt{2} \, \mu C = -18\sqrt{2} \times 10^{-6} \, C$. Its position is $x_2 = 4$ m, so its position vector is $\vec{r}_{q2} = (4, 0)$ m.
• The point P where the electric field is to be calculated is $(1, 3)$ m. Its position vector is $\vec{r}_P = (1, 3)$ m.
• The electrostatic constant $k = \frac{1}{4\pi\epsilon_0} = 9 \times 10^9 \, Nm^2C^{-2}$.

Step-by-Step Solution for Electric Field Calculation

We will calculate the electric field contribution from each charge separately and then add them vectorially.

Field Contribution from Charge $q_1$

The electric field $\vec{E}_1$ at point P due to charge $q_1$ is given by the formula $\vec{E}_1 = k \frac{q_1}{r_1^2} \hat{r}_1$, where $r_1$ is the distance from $q_1$ to P, and $\hat{r}_1$ is the unit vector pointing from $q_1$ to P.

• Position Vector from $q_1$ to P: $\vec{r}_1 = \vec{r}_P - \vec{r}_{q1} = (1, 3) - (0, 0) = (1, 3)$ m. In unit vector notation, $\vec{r}_1 = (1)\hat{i} + (3)\hat{j}$ m.
• Distance $r_1$: $r_1 = |\vec{r}_1| = \sqrt{1^2 + 3^2} = \sqrt{1 + 9} = \sqrt{10}$ m.
• Square of Distance $r_1^2$: $r_1^2 = (\sqrt{10})^2 = 10$ m$^2$.
• Unit Vector $\hat{r}_1$: $\hat{r}_1 = \frac{\vec{r}_1}{r_1} = \frac{(1)\hat{i} + (3)\hat{j}}{\sqrt{10}} = \frac{1}{\sqrt{10}}\hat{i} + \frac{3}{\sqrt{10}}\hat{j}$.
• Calculating $\vec{E}_1$: $\vec{E}_1 = \left( 9 \times 10^9 \frac{Nm^2}{C^2} \right) \times \frac{\sqrt{10} \times 10^{-6} \, C}{10 \, m^2} \times \left( \frac{1}{\sqrt{10}}\hat{i} + \frac{3}{\sqrt{10}}\hat{j} \right)$ $\vec{E}_1 = \left( 9 \times 10^3 \frac{Nm}{C} \right) \times \frac{1}{10} \times \left( \frac{\sqrt{10}}{\sqrt{10}}\hat{i} + \frac{3\sqrt{10}}{\sqrt{10}}\hat{j} \right)$ $\vec{E}_1 = \left( 9 \times 10^2 \frac{Nm}{C} \right) \times (1\hat{i} + 3\hat{j})$ $\vec{E}_1 = (900 \hat{i} + 2700 \hat{j})$ V/m.

Field Contribution from Charge $q_2$

Similarly, the electric field $\vec{E}_2$ at point P due to charge $q_2$ is given by $\vec{E}_2 = k \frac{q_2}{r_2^2} \hat{r}_2$, where $r_2$ is the distance from $q_2$ to P, and $\hat{r}_2$ is the unit vector pointing from $q_2$ to P.

• Position Vector from $q_2$ to P: $\vec{r}_2 = \vec{r}_P - \vec{r}_{q2} = (1, 3) - (4, 0) = (-3, 3)$ m. In unit vector notation, $\vec{r}_2 = (-3)\hat{i} + (3)\hat{j}$ m.
• Distance $r_2$: $r_2 = |\vec{r}_2| = \sqrt{(-3)^2 + 3^2} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2}$ m.
• Square of Distance $r_2^2$: $r_2^2 = (\sqrt{18})^2 = 18$ m$^2$.
• Unit Vector $\hat{r}_2$: $\hat{r}_2 = \frac{\vec{r}_2}{r_2} = \frac{(-3)\hat{i} + (3)\hat{j}}{3\sqrt{2}} = \frac{-1}{\sqrt{2}}\hat{i} + \frac{1}{\sqrt{2}}\hat{j}$.
• Calculating $\vec{E}_2$: $\vec{E}_2 = \left( 9 \times 10^9 \frac{Nm^2}{C^2} \right) \times \frac{-18\sqrt{2} \times 10^{-6} \, C}{18 \, m^2} \times \left( \frac{-1}{\sqrt{2}}\hat{i} + \frac{1}{\sqrt{2}}\hat{j} \right)$ $\vec{E}_2 = \left( 9 \times 10^3 \frac{Nm}{C} \right) \times (-\sqrt{2}) \times \left( \frac{-1}{\sqrt{2}}\hat{i} + \frac{1}{\sqrt{2}}\hat{j} \right)$ $\vec{E}_2 = \left( -9\sqrt{2} \times 10^3 \frac{Nm}{C} \right) \times \left( \frac{-1}{\sqrt{2}}\hat{i} + \frac{1}{\sqrt{2}}\hat{j} \right)$ $\vec{E}_2 = \left( -9\sqrt{2} \times 10^3 \times \frac{-1}{\sqrt{2}} \right) \hat{i} + \left( -9\sqrt{2} \times 10^3 \times \frac{1}{\sqrt{2}} \right) \hat{j}$ $\vec{E}_2 = (9 \times 10^3) \hat{i} - (9 \times 10^3) \hat{j}$ $\vec{E}_2 = (9000 \hat{i} - 9000 \hat{j})$ V/m.

Total Electric Field Vector Sum

According to the principle of superposition, the total electric field $\vec{E}_{total}$ at point P is the vector sum of $\vec{E}_1$ and $\vec{E}_2$. $ \vec{E}_{total} = \vec{E}_1 + \vec{E}_2 $

• Adding the components: $\vec{E}_{total} = (900 \hat{i} + 2700 \hat{j}) \, V/m + (9000 \hat{i} - 9000 \hat{j}) \, V/m$ $\vec{E}_{total} = (900 + 9000) \hat{i} + (2700 - 9000) \hat{j}$ V/m $\vec{E}_{total} = (9900 \hat{i} - 6300 \hat{j})$ V/m.

Final Result and Option Comparison

The calculated total electric field is $(9900 \hat{i} - 6300 \hat{j})$ V/m. To compare this with the given options, we can factor out $10^2$:

• $\vec{E}_{total} = (99 \times 10^2 \hat{i} - 63 \times 10^2 \hat{j})$ V/m
• $\vec{E}_{total} = (99\hat{i} - 63\hat{j}) \times 10^2$ V/m.

This result matches the first option.