Question

Two point charges \( q_1 = 16 \, \mu C \) and \( q_2 = 1 \, \mu C \) are placed at points \( \vec{r}_1 = (3 \, \text{m}) \hat{i}\) and \( \vec{r}_2 = (4 \, \text{m}) \hat{j} \). Find the net electric field \( \vec{E} \) at point \( \vec{r} = (3 \, \text{m}) \hat{i} + (4 \, \text{m}) \hat{j} \).

Hint:

When calculating the net electric field from multiple charges, calculate the electric field due to each charge separately and then combine the results vectorially, taking care of the direction.

Answer 1

Net Electric Field at a Point Due to Two Point Charges 

We are given two point charges and need to find the net electric field at a point \( \vec{r} = 3 \hat{i} + 4 \hat{j} \) due to the two charges.

The electric field due to a point charge is given by Coulomb's law:

\[ \vec{E} = \frac{1}{4 \pi \varepsilon_0} \cdot \frac{q}{r^2} \hat{r} \]

where:

• \( q \) is the charge,
• \( r \) is the distance from the charge to the point of interest,
• \( \hat{r} \) is the unit vector pointing from the charge to the point,
• \( \varepsilon_0 \) is the permittivity of free space \( \varepsilon_0 = 8.85 \times 10^{-12} \, \text{C}^2/\text{N} \cdot \text{m}^2 \).

Electric Field Due to \( q_1 \) at Point \( \vec{r} \):

First, calculate the distance \( r_1 \) between \( q_1 \) and point \( \vec{r} \). Since \( q_1 \) is at \( (3, 0) \) and point \( \vec{r} \) is at \( (3, 4) \):

\[ r_1 = \sqrt{(3 - 3)^2 + (4 - 0)^2} = \sqrt{16} = 4 \, \text{m} \]

Next, find the unit vector \( \hat{r_1} \):

\[ \hat{r_1} = \frac{\vec{r} - \vec{r_1}}{r_1} = \frac{(3 \hat{i} + 4 \hat{j}) - (3 \hat{i})}{4} = \hat{j} \]

Now, using Coulomb's law for \( q_1 \):

\[ \vec{E_1} = \frac{1}{4 \pi \varepsilon_0} \cdot \frac{q_1}{r_1^2} \hat{r_1} = \frac{1}{4 \pi \varepsilon_0} \cdot \frac{16 \times 10^{-6}}{(4)^2} \hat{j} = \frac{16 \times 10^{-6}}{16 \pi \varepsilon_0} \hat{j} \]

Electric Field Due to \( q_2 \) at Point \( \vec{r} \):

Similarly, calculate the distance \( r_2 \) between \( q_2 \) and point \( \vec{r} \):

\[ r_2 = \sqrt{(3 - 0)^2 + (4 - 4)^2} = \sqrt{9} = 3 \, \text{m} \]

Find the unit vector \( \hat{r_2} \):

\[ \hat{r_2} = \frac{\vec{r} - \vec{r_2}}{r_2} = \frac{(3 \hat{i} + 4 \hat{j}) - (0 \hat{i} + 4 \hat{j})}{3} = \frac{3 \hat{i}}{3} = \hat{i} \]

Now, using Coulomb's law for \( q_2 \):

\[ \vec{E_2} = \frac{1}{4 \pi \varepsilon_0} \cdot \frac{q_2}{r_2^2} \hat{r_2} = \frac{1}{4 \pi \varepsilon_0} \cdot \frac{1 \times 10^{-6}}{(3)^2} \hat{i} = \frac{1 \times 10^{-6}}{9 \pi \varepsilon_0} \hat{i} \]

Net Electric Field at Point \( \vec{r} \):

The total electric field \( \vec{E} \) at point \( \vec{r} \) is the vector sum of \( \vec{E_1} \) and \( \vec{E_2} \):

\[ \vec{E} = \vec{E_1} + \vec{E_2} = \frac{16 \times 10^{-6}}{16 \pi \varepsilon_0} \hat{j} + \frac{1 \times 10^{-6}}{9 \pi \varepsilon_0} \hat{i} \]

Thus, the net electric field at point \( \vec{r} \) is:

\[ \vec{E} = \frac{1 \times 10^{-6}}{9 \pi \varepsilon_0} \hat{i} + \frac{16 \times 10^{-6}}{16 \pi \varepsilon_0} \hat{j} \]