In a vacuum, two point charges with magnitudes of +1.8 nC and -1.8 nC are separated by 6 mm along the x-axis. At the halfway point of the separation distance, the electric field is _____.
3.6 × 106 NC-1
To determine the electric field at the halfway point between two point charges, we need to consider the electric field contributed by each charge separately and then add them vectorially.
Here's a breakdown of the given information:
- First point charge, \(q_1 = +1.8 \, \text{nC} = +1.8 \times 10^{-9} \, \text{C}\)
- Second point charge, \(q_2 = -1.8 \, \text{nC} = -1.8 \times 10^{-9} \, \text{C}\)
- Separation distance between the charges, \(d = 6 \, \text{mm} = 6 \times 10^{-3} \, \text{m}\)
The problem asks for the electric field at the halfway point of the separation distance. This means the point of interest is \(r = d/2\) from each charge.
So, the distance from each charge to the halfway point is:
\(r = \frac{6 \, \text{mm}}{2} = 3 \, \text{mm} = 3 \times 10^{-3} \, \text{m}\)
We will use Coulomb's Law for the electric field due to a point charge, which is given by:
\(E = \frac{k|q|}{r^2}\)
Where:
- \(E\) is the magnitude of the electric field
- \(k\) is Coulomb's constant, approximately \(9 \times 10^9 \, \text{N} \cdot \text{m}^2/\text{C}^2\)
- \(|q|\) is the magnitude of the point charge
- \(r\) is the distance from the charge to the point where the field is being calculated
Electric Field from Positive Charge
Let's calculate the electric field \(E_1\) due to the positive charge \(q_1 = +1.8 \times 10^{-9} \, \text{C}\) at the halfway point.
\(E_1 = \frac{(9 \times 10^9 \, \text{N} \cdot \text{m}^2/\text{C}^2) \times |+1.8 \times 10^{-9} \, \text{C}|}{(3 \times 10^{-3} \, \text{m})^2}\)
\(E_1 = \frac{9 \times 10^9 \times 1.8 \times 10^{-9}}{ (3 \times 10^{-3})^2}\)
\(E_1 = \frac{16.2}{9 \times 10^{-6}}\)
\(E_1 = 1.8 \times 10^6 \, \text{NC}^{-1}\)
Since the charge \(q_1\) is positive, the electric field \(E_1\) at the halfway point will point away from \(q_1\). If we imagine \(q_1\) is at the left and \(q_2\) is at the right along the x-axis, the field \(E_1\) will point to the right (positive x-direction).
Electric Field from Negative Charge
Next, let's calculate the electric field \(E_2\) due to the negative charge \(q_2 = -1.8 \times 10^{-9} \, \text{C}\) at the halfway point.
\(E_2 = \frac{(9 \times 10^9 \, \text{N} \cdot \text{m}^2/\text{C}^2) \times |-1.8 \times 10^{-9} \, \text{C}|}{(3 \times 10^{-3} \, \text{m})^2}\)
\(E_2 = \frac{9 \times 10^9 \times 1.8 \times 10^{-9}}{ (3 \times 10^{-3})^2}\)
\(E_2 = \frac{16.2}{9 \times 10^{-6}}\)
\(E_2 = 1.8 \times 10^6 \, \text{NC}^{-1}\)
Since the charge \(q_2\) is negative, the electric field \(E_2\) at the halfway point will point towards \(q_2\). Given our setup (positive charge at left, negative charge at right, and halfway point in between), the field \(E_2\) will also point towards \(q_2\) (i.e., in the positive x-direction, to the right).
Total Electric Field
Since both electric fields, \(E_1\) and \(E_2\), are in the same direction (both pointing to the right), we can simply add their magnitudes to find the total electric field \(E_{total}\) at the halfway point.
\(E_{total} = E_1 + E_2\)
\(E_{total} = (1.8 \times 10^6 \, \text{NC}^{-1}) + (1.8 \times 10^6 \, \text{NC}^{-1})\)
\(E_{total} = (1.8 + 1.8) \times 10^6 \, \text{NC}^{-1}\)
\(E_{total} = 3.6 \times 10^6 \, \text{NC}^{-1}\)
The total electric field at the halfway point is \(3.6 \times 10^6 \, \text{NC}^{-1}\).
| Parameter | Value |
|---|---|
| Charge \(q_1\) | \(+1.8 \times 10^{-9}\) C |
| Charge \(q_2\) | \(-1.8 \times 10^{-9}\) C |
| Total separation \(d\) | \(6 \times 10^{-3}\) m |
| Distance to midpoint \(r\) | \(3 \times 10^{-3}\) m |
| Coulomb's constant \(k\) | \(9 \times 10^9\) N·m2/C2 |
| Electric field \(E_1\) (from \(q_1\)) | \(1.8 \times 10^6\) NC-1 (Right) |
| Electric field \(E_2\) (from \(q_2\)) | \(1.8 \times 10^6\) NC-1 (Right) |
| Total Electric Field \(E_{total}\) | \(3.6 \times 10^6\) NC-1 (Right) |
Important Questions from Electric Fields and Gauss' Law
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If a free electron moves through a potential difference of 1 kV, then the energy gained by the electron is given by
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