Question
How do I calculate the electric field at the origin for two point charges, 3.0×10^-6 C and -2.0×10^-6 C, located at x = -2.0 cm and x = 4.0 cm?
Answers
GPT-4o mini
To calculate the electric field at the origin due to two point charges, we can use the formula for the electric field \( E \) created by a point charge, which is given by:
\[
E = \frac{k \cdot |q|}{r^2}
\]
where:
- \( E \) is the electric field,
- \( k \) is Coulomb's constant (\( k \approx 8.99 \times 10^9 \, \text{N m}^2/\text{C}^2 \)),
- \( q \) is the charge,
- \( r \) is the distance from the charge to the point where the electric field is being calculated.
### Step 1: Determine the distances from the charges to the origin
1. For the charge \( q_1 = 3.0 \times 10^{-6} \, C \) located at \( x = -2.0 \, \text{cm} = -0.02 \, \text{m} \):
- The distance \( r_1 \) from the charge to the origin (0,0) is:
\[
r_1 = |-0.02 \, \text{m}| = 0.02 \, \text{m}
\]
2. For the charge \( q_2 = -2.0 \times 10^{-6} \, C \) located at \( x = 4.0 \, \text{cm} = 0.04 \, \text{m} \):
- The distance \( r_2 \) from the charge to the origin is:
\[
r_2 = |0.04 \, \text{m}| = 0.04 \, \text{m}
\]
### Step 2: Calculate the electric fields due to each charge
1. **Electric field due to \( q_1 \)**:
\[
E_1 = \frac{k \cdot |q_1|}{r_1^2}
\]
Substituting the values:
\[
E_1 = \frac{(8.99 \times 10^9) \cdot (3.0 \times 10^{-6})}{(0.02)^2}
\]
\[
E_1 = \frac{(8.99 \times 10^9) \cdot (3.0 \times 10^{-6})}{0.0004} \approx 6.7425 \times 10^{5} \, \text{N/C}
\]
The direction of \( E_1 \) is away from the positive charge, towards the right (in the positive x direction).
2. **Electric field due to \( q_2 \)**:
\[
E_2 = \frac{k \cdot |q_2|}{r_2^2}
\]
Substituting the values:
\[
E_2 = \frac{(8.99 \times 10^9) \cdot (2.0 \times 10^{-6})}{(0.04)^2}
\]
\[
E_2 = \frac{(8.99 \times 10^9) \cdot (2.0 \times 10^{-6})}{0.0016} \approx 1.12375 \times 10^{5} \, \text{N/C}
\]
The direction of \( E_2 \) is towards the negative charge, which is also to the left (in the negative x direction).
### Step 3: Determine the net electric field at the origin
Since \( E_1 \) is in the positive x direction and \( E_2 \) is in the negative x direction, we can write:
\[
E_{\text{net}} = E_1 - E_2
\]
Substituting the values we calculated:
\[
E_{\text{net}} = 6.7425 \times 10^5 - 1.12375 \times 10^5 = 5.61875 \times 10^5 \, \text{N/C}
\]
### Conclusion
The net electric field at the origin is approximately:
\[
E_{\text{net}} \approx 5.6 \times 10^5 \, \text{N/C} \, \text{(positive x direction)}
\]
\[
E = \frac{k \cdot |q|}{r^2}
\]
where:
- \( E \) is the electric field,
- \( k \) is Coulomb's constant (\( k \approx 8.99 \times 10^9 \, \text{N m}^2/\text{C}^2 \)),
- \( q \) is the charge,
- \( r \) is the distance from the charge to the point where the electric field is being calculated.
### Step 1: Determine the distances from the charges to the origin
1. For the charge \( q_1 = 3.0 \times 10^{-6} \, C \) located at \( x = -2.0 \, \text{cm} = -0.02 \, \text{m} \):
- The distance \( r_1 \) from the charge to the origin (0,0) is:
\[
r_1 = |-0.02 \, \text{m}| = 0.02 \, \text{m}
\]
2. For the charge \( q_2 = -2.0 \times 10^{-6} \, C \) located at \( x = 4.0 \, \text{cm} = 0.04 \, \text{m} \):
- The distance \( r_2 \) from the charge to the origin is:
\[
r_2 = |0.04 \, \text{m}| = 0.04 \, \text{m}
\]
### Step 2: Calculate the electric fields due to each charge
1. **Electric field due to \( q_1 \)**:
\[
E_1 = \frac{k \cdot |q_1|}{r_1^2}
\]
Substituting the values:
\[
E_1 = \frac{(8.99 \times 10^9) \cdot (3.0 \times 10^{-6})}{(0.02)^2}
\]
\[
E_1 = \frac{(8.99 \times 10^9) \cdot (3.0 \times 10^{-6})}{0.0004} \approx 6.7425 \times 10^{5} \, \text{N/C}
\]
The direction of \( E_1 \) is away from the positive charge, towards the right (in the positive x direction).
2. **Electric field due to \( q_2 \)**:
\[
E_2 = \frac{k \cdot |q_2|}{r_2^2}
\]
Substituting the values:
\[
E_2 = \frac{(8.99 \times 10^9) \cdot (2.0 \times 10^{-6})}{(0.04)^2}
\]
\[
E_2 = \frac{(8.99 \times 10^9) \cdot (2.0 \times 10^{-6})}{0.0016} \approx 1.12375 \times 10^{5} \, \text{N/C}
\]
The direction of \( E_2 \) is towards the negative charge, which is also to the left (in the negative x direction).
### Step 3: Determine the net electric field at the origin
Since \( E_1 \) is in the positive x direction and \( E_2 \) is in the negative x direction, we can write:
\[
E_{\text{net}} = E_1 - E_2
\]
Substituting the values we calculated:
\[
E_{\text{net}} = 6.7425 \times 10^5 - 1.12375 \times 10^5 = 5.61875 \times 10^5 \, \text{N/C}
\]
### Conclusion
The net electric field at the origin is approximately:
\[
E_{\text{net}} \approx 5.6 \times 10^5 \, \text{N/C} \, \text{(positive x direction)}
\]