Answer :
Sure! To find the electric field at point [tex]\( P \)[/tex] from the given charges, we can follow a step-by-step approach. Let's first visualize the scenario with a diagram.
```markdown
Diagram:
Y-axis
|
P (0, 3m)
| q2(4m, 0)
| /
| /
| /
| /
| /
| /
| /
| /
| / q1(0, 0)
| /________________________________________ X-axis
```
We'll calculate the electric field contributions from [tex]\( q_1 \)[/tex] and [tex]\( q_2 \)[/tex] at point [tex]\( P \)[/tex].
1. Constants and positions:
- Charge [tex]\( q_1 = +10 \, \text{nC} = 10 \times 10^{-9} \, \text{C}\)[/tex],
- Charge [tex]\( q_2 = +15 \, \text{nC} = 15 \times 10^{-9} \, \text{C}\)[/tex],
- Electrostatic constant [tex]\( k = 8.99 \times 10^9 \, \text{N} \cdot \text{m}^2 / \text{C}^2 \)[/tex],
- Position of [tex]\( q_1 \)[/tex]: [tex]\((0, 0)\)[/tex],
- Position of [tex]\( q_2 \)[/tex]: [tex]\((4 \, \text{m}, 0)\)[/tex],
- Position of point [tex]\( P \)[/tex]: [tex]\((0, 3 \, \text{m})\)[/tex].
2. Calculate the distances from [tex]\( q_1 \)[/tex] and [tex]\( q_2 \)[/tex] to point [tex]\( P \)[/tex]:
[tex]\[ r_1 = \sqrt{(0 - 0)^2 + (3 - 0)^2} = 3 \, \text{m} \][/tex]
[tex]\[ r_2 = \sqrt{(0 - 4)^2 + (3 - 0)^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \, \text{m} \][/tex]
3. Calculate the magnitudes of the electric field components from each charge:
[tex]\[ E_1 = \frac{k \times q_1}{r_1^2} = \frac{8.99 \times 10^9 \times 10 \times 10^{-9}}{3^2} = 9.988888888888889 \, \text{N/C} \][/tex]
[tex]\[ E_2 = \frac{k \times q_2}{r_2^2} = \frac{8.99 \times 10^9 \times 15 \times 10^{-9}}{5^2} = 5.394 \, \text{N/C} \][/tex]
4. Breakdown the electric field from [tex]\( q_1 \)[/tex] and [tex]\( q_2 \)[/tex] into components:
- For [tex]\( q_1 \)[/tex], the electric field [tex]\( E_1 \)[/tex] is entirely in the positive y-direction since [tex]\( P \)[/tex] is directly above [tex]\( q_1 \)[/tex]:
[tex]\[ E_{1x} = 0 \][/tex]
[tex]\[ E_{1y} = 9.988888888888889 \, \text{N/C} \][/tex]
- For [tex]\( q_2 \)[/tex], we need to break down [tex]\( E_2 \)[/tex] into x and y components:
[tex]\[ E_{2x} = - E_2 \times \frac{4}{5} = - 5.394 \times \frac{4}{5} = 4.3152 \, \text{N/C} \][/tex]
[tex]\[ E_{2y} = - E_2 \times \frac{3}{5} = - 5.394 \times \frac{3}{5} = -3.2364000000000006 \, \text{N/C} \][/tex]
5. Sum the components of the electric fields at point [tex]\( P \)[/tex]:
- X-component:
[tex]\[ E_x = E_{1x} + E_{2x} = 0 + 4.3152 = 4.3152 \, \text{N/C} \][/tex]
- Y-component:
[tex]\[ E_y = E_{1y} + E_{2y} = 9.988888888888889 + (-3.2364000000000006) = 6.752488888888888 \, \text{N/C} \][/tex]
6. Calculate the resultant electric field at point [tex]\( P \)[/tex]:
[tex]\[ E_{\text{total}} = \sqrt{E_x^2 + E_y^2} = \sqrt{(4.3152)^2 + (6.752488888888888)^2} \approx 8.01355459422146 \, \text{N/C} \][/tex]
So, the magnitudes of the electric fields from each charge and the resultant field at point [tex]\( P \)[/tex] are as follows:
- Field from [tex]\( q_1 \)[/tex]: [tex]\( 9.988888888888889 \, \text{N/C} \)[/tex],
- Field from [tex]\( q_2 \)[/tex]: [tex]\( 5.394 \, \text{N/C} \)[/tex],
- Resultant electric field at [tex]\( P \)[/tex] [tex]\( (4.3152 \, \text{N/C}, 6.752488888888888 \, \text{N/C}) \)[/tex],
- Total magnitude of the electric field at [tex]\( P \)[/tex]: [tex]\( 8.01355459422146 \, \text{N/C} \)[/tex].
This gives us a detailed, step-by-step solution for determining the electric field at point [tex]\( P \)[/tex], considering both the contributions from [tex]\( q_1 \)[/tex] and [tex]\( q_2 \)[/tex].
```markdown
Diagram:
Y-axis
|
P (0, 3m)
| q2(4m, 0)
| /
| /
| /
| /
| /
| /
| /
| /
| / q1(0, 0)
| /________________________________________ X-axis
```
We'll calculate the electric field contributions from [tex]\( q_1 \)[/tex] and [tex]\( q_2 \)[/tex] at point [tex]\( P \)[/tex].
1. Constants and positions:
- Charge [tex]\( q_1 = +10 \, \text{nC} = 10 \times 10^{-9} \, \text{C}\)[/tex],
- Charge [tex]\( q_2 = +15 \, \text{nC} = 15 \times 10^{-9} \, \text{C}\)[/tex],
- Electrostatic constant [tex]\( k = 8.99 \times 10^9 \, \text{N} \cdot \text{m}^2 / \text{C}^2 \)[/tex],
- Position of [tex]\( q_1 \)[/tex]: [tex]\((0, 0)\)[/tex],
- Position of [tex]\( q_2 \)[/tex]: [tex]\((4 \, \text{m}, 0)\)[/tex],
- Position of point [tex]\( P \)[/tex]: [tex]\((0, 3 \, \text{m})\)[/tex].
2. Calculate the distances from [tex]\( q_1 \)[/tex] and [tex]\( q_2 \)[/tex] to point [tex]\( P \)[/tex]:
[tex]\[ r_1 = \sqrt{(0 - 0)^2 + (3 - 0)^2} = 3 \, \text{m} \][/tex]
[tex]\[ r_2 = \sqrt{(0 - 4)^2 + (3 - 0)^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \, \text{m} \][/tex]
3. Calculate the magnitudes of the electric field components from each charge:
[tex]\[ E_1 = \frac{k \times q_1}{r_1^2} = \frac{8.99 \times 10^9 \times 10 \times 10^{-9}}{3^2} = 9.988888888888889 \, \text{N/C} \][/tex]
[tex]\[ E_2 = \frac{k \times q_2}{r_2^2} = \frac{8.99 \times 10^9 \times 15 \times 10^{-9}}{5^2} = 5.394 \, \text{N/C} \][/tex]
4. Breakdown the electric field from [tex]\( q_1 \)[/tex] and [tex]\( q_2 \)[/tex] into components:
- For [tex]\( q_1 \)[/tex], the electric field [tex]\( E_1 \)[/tex] is entirely in the positive y-direction since [tex]\( P \)[/tex] is directly above [tex]\( q_1 \)[/tex]:
[tex]\[ E_{1x} = 0 \][/tex]
[tex]\[ E_{1y} = 9.988888888888889 \, \text{N/C} \][/tex]
- For [tex]\( q_2 \)[/tex], we need to break down [tex]\( E_2 \)[/tex] into x and y components:
[tex]\[ E_{2x} = - E_2 \times \frac{4}{5} = - 5.394 \times \frac{4}{5} = 4.3152 \, \text{N/C} \][/tex]
[tex]\[ E_{2y} = - E_2 \times \frac{3}{5} = - 5.394 \times \frac{3}{5} = -3.2364000000000006 \, \text{N/C} \][/tex]
5. Sum the components of the electric fields at point [tex]\( P \)[/tex]:
- X-component:
[tex]\[ E_x = E_{1x} + E_{2x} = 0 + 4.3152 = 4.3152 \, \text{N/C} \][/tex]
- Y-component:
[tex]\[ E_y = E_{1y} + E_{2y} = 9.988888888888889 + (-3.2364000000000006) = 6.752488888888888 \, \text{N/C} \][/tex]
6. Calculate the resultant electric field at point [tex]\( P \)[/tex]:
[tex]\[ E_{\text{total}} = \sqrt{E_x^2 + E_y^2} = \sqrt{(4.3152)^2 + (6.752488888888888)^2} \approx 8.01355459422146 \, \text{N/C} \][/tex]
So, the magnitudes of the electric fields from each charge and the resultant field at point [tex]\( P \)[/tex] are as follows:
- Field from [tex]\( q_1 \)[/tex]: [tex]\( 9.988888888888889 \, \text{N/C} \)[/tex],
- Field from [tex]\( q_2 \)[/tex]: [tex]\( 5.394 \, \text{N/C} \)[/tex],
- Resultant electric field at [tex]\( P \)[/tex] [tex]\( (4.3152 \, \text{N/C}, 6.752488888888888 \, \text{N/C}) \)[/tex],
- Total magnitude of the electric field at [tex]\( P \)[/tex]: [tex]\( 8.01355459422146 \, \text{N/C} \)[/tex].
This gives us a detailed, step-by-step solution for determining the electric field at point [tex]\( P \)[/tex], considering both the contributions from [tex]\( q_1 \)[/tex] and [tex]\( q_2 \)[/tex].
