Answer :
We start by recalling that the electric field produced by a point charge is given by
[tex]$$
E = \frac{k q}{r^2},
$$[/tex]
where
- [tex]$k = 8.9875517923\times10^9\ \text{N·m}^2/\text{C}^2$[/tex] is Coulomb’s constant,
- [tex]$q$[/tex] is the charge,
- [tex]$r$[/tex] is the distance from the charge to the point of interest.
Here we have two charges:
- [tex]$q_1 = 46.5\,\mu\text{C}$[/tex] located above point [tex]$P$[/tex] at a distance [tex]$r_1 = 10.1\,\text{m}$[/tex],
- [tex]$q_2 = 46.5\,\mu\text{C}$[/tex] located to the right of point [tex]$P$[/tex] at a distance [tex]$r_2 = 5.8\,\text{m}$[/tex].
Step 1. Calculate the electric field magnitudes due to each charge.
For charge [tex]$q_1$[/tex], the magnitude of the electric field at [tex]$P$[/tex] is:
[tex]$$
E_1 = \frac{kq_1}{r_1^2} = \frac{8.9875517923\times10^9 \times 46.5\times10^{-6}}{(10.1)^2}.
$$[/tex]
Evaluating this gives
[tex]$$
E_1 \approx 4096.86\ \text{N/C}.
$$[/tex]
For charge [tex]$q_2$[/tex], the magnitude is:
[tex]$$
E_2 = \frac{kq_2}{r_2^2} = \frac{8.9875517923\times10^9 \times 46.5\times10^{-6}}{(5.8)^2}.
$$[/tex]
Evaluating this gives
[tex]$$
E_2 \approx 12423.34\ \text{N/C}.
$$[/tex]
Step 2. Determine the direction of each electric field.
Because both charges are positive, the electric field produced by each charge points radially outward from that charge.
- For [tex]$q_1$[/tex] located above point [tex]$P$[/tex], the electric field at [tex]$P$[/tex] points away from [tex]$q_1$[/tex], i.e., vertically downward (in the negative [tex]$y$[/tex]-direction).
- For [tex]$q_2$[/tex] located to the right of point [tex]$P$[/tex], the electric field at [tex]$P$[/tex] points away from [tex]$q_2$[/tex], i.e., horizontally to the left (in the negative [tex]$x$[/tex]-direction).
In unit vector notation, this means:
- The electric field from [tex]$q_1$[/tex] is [tex]$$\vec{E}_1 = -E_1\,\hat{j}.$$[/tex]
- The electric field from [tex]$q_2$[/tex] is [tex]$$\vec{E}_2 = -E_2\,\hat{i}.$$[/tex]
Step 3. Combine the contributions to get the net electric field at point [tex]$P$[/tex].
Since the fields are perpendicular, we add the components:
- The net [tex]$x$[/tex]-component is just from [tex]$q_2$[/tex]:
[tex]$$
E_{\text{net},x} = -12423.34\ \text{N/C}.
$$[/tex]
- The net [tex]$y$[/tex]-component is just from [tex]$q_1$[/tex]:
[tex]$$
E_{\text{net},y} = -4096.86\ \text{N/C}.
$$[/tex]
Thus, the net electric field at [tex]$P$[/tex] in unit vector notation is:
[tex]$$
\vec{E}_{\text{net}} = -12423.34\,\hat{i} - 4096.86\,\hat{j}\ \text{N/C}.
$$[/tex]
This is your final answer.
[tex]$$
E = \frac{k q}{r^2},
$$[/tex]
where
- [tex]$k = 8.9875517923\times10^9\ \text{N·m}^2/\text{C}^2$[/tex] is Coulomb’s constant,
- [tex]$q$[/tex] is the charge,
- [tex]$r$[/tex] is the distance from the charge to the point of interest.
Here we have two charges:
- [tex]$q_1 = 46.5\,\mu\text{C}$[/tex] located above point [tex]$P$[/tex] at a distance [tex]$r_1 = 10.1\,\text{m}$[/tex],
- [tex]$q_2 = 46.5\,\mu\text{C}$[/tex] located to the right of point [tex]$P$[/tex] at a distance [tex]$r_2 = 5.8\,\text{m}$[/tex].
Step 1. Calculate the electric field magnitudes due to each charge.
For charge [tex]$q_1$[/tex], the magnitude of the electric field at [tex]$P$[/tex] is:
[tex]$$
E_1 = \frac{kq_1}{r_1^2} = \frac{8.9875517923\times10^9 \times 46.5\times10^{-6}}{(10.1)^2}.
$$[/tex]
Evaluating this gives
[tex]$$
E_1 \approx 4096.86\ \text{N/C}.
$$[/tex]
For charge [tex]$q_2$[/tex], the magnitude is:
[tex]$$
E_2 = \frac{kq_2}{r_2^2} = \frac{8.9875517923\times10^9 \times 46.5\times10^{-6}}{(5.8)^2}.
$$[/tex]
Evaluating this gives
[tex]$$
E_2 \approx 12423.34\ \text{N/C}.
$$[/tex]
Step 2. Determine the direction of each electric field.
Because both charges are positive, the electric field produced by each charge points radially outward from that charge.
- For [tex]$q_1$[/tex] located above point [tex]$P$[/tex], the electric field at [tex]$P$[/tex] points away from [tex]$q_1$[/tex], i.e., vertically downward (in the negative [tex]$y$[/tex]-direction).
- For [tex]$q_2$[/tex] located to the right of point [tex]$P$[/tex], the electric field at [tex]$P$[/tex] points away from [tex]$q_2$[/tex], i.e., horizontally to the left (in the negative [tex]$x$[/tex]-direction).
In unit vector notation, this means:
- The electric field from [tex]$q_1$[/tex] is [tex]$$\vec{E}_1 = -E_1\,\hat{j}.$$[/tex]
- The electric field from [tex]$q_2$[/tex] is [tex]$$\vec{E}_2 = -E_2\,\hat{i}.$$[/tex]
Step 3. Combine the contributions to get the net electric field at point [tex]$P$[/tex].
Since the fields are perpendicular, we add the components:
- The net [tex]$x$[/tex]-component is just from [tex]$q_2$[/tex]:
[tex]$$
E_{\text{net},x} = -12423.34\ \text{N/C}.
$$[/tex]
- The net [tex]$y$[/tex]-component is just from [tex]$q_1$[/tex]:
[tex]$$
E_{\text{net},y} = -4096.86\ \text{N/C}.
$$[/tex]
Thus, the net electric field at [tex]$P$[/tex] in unit vector notation is:
[tex]$$
\vec{E}_{\text{net}} = -12423.34\,\hat{i} - 4096.86\,\hat{j}\ \text{N/C}.
$$[/tex]
This is your final answer.