Step 1: Understanding the Concept:
The electric field (\(E\)) at a point in space is defined as the force per unit charge exerted on a positive test charge placed at that point.
When dealing with multiple point charges, we apply the Principle of Superposition.
This principle states that the net electric field at any point is the vector sum of the individual electric fields produced by each charge.
Crucially, the electric field is a vector. For a positive charge, the field vector points radially outward (away from the charge).
For a negative charge, the field vector points radially inward (towards the charge).
At the midpoint of two charges, the distance to each charge is exactly half the total separation distance.
Step 2: Key Formula or Approach:
The magnitude of the electric field produced by a point charge \( q \) at a distance \( r \) is given by Coulomb's Law variant:
\[ E = \frac{k|q|}{r^{2}} \]
Where \( k \) is the electrostatic constant, approximately \( 9 \times 10^{9} \text{ N m}^{2} \text{ C}^{-2} \).
The total field is:
\[ \vec{E}_{\text{net}} = \vec{E}_{1} + \vec{E}_{2} \]
Step 3: Detailed Explanation:
Let's analyze the setup:
Charge 1 (\( q_{1} \)) = \( +2 \mu\text{C} = +2 \times 10^{-6} \text{ C} \).
Charge 2 (\( q_{2} \)) = \( -2 \mu\text{C} = -2 \times 10^{-6} \text{ C} \).
Separation distance \( d = 2 \text{ m} \).
The midpoint is located at distance \( r = \frac{d}{2} = 1 \text{ m} \) from both charges.
1. Direction Analysis:
- Imagine a small positive test charge at the midpoint.
- The positive charge (\( q_{1} \)) on the left will repel the test charge to the right. So, \( \vec{E}_{1} \) points right.
- The negative charge (\( q_{2} \)) on the right will attract the test charge towards itself. So, \( \vec{E}_{2} \) also points right.
- Since both vectors point in the same direction, the net field magnitude is the sum of the individual magnitudes: \( E_{\text{net}} = E_{1} + E_{2} \).
2. Magnitude Calculations:
\[ E_{1} = \frac{(9 \times 10^{9})(2 \times 10^{-6})}{1^{2}} = 18 \times 10^{3} \text{ N/C} \]
\[ E_{2} = \frac{(9 \times 10^{9})(2 \times 10^{-6})}{1^{2}} = 18 \times 10^{3} \text{ N/C} \]
3. Final Summation:
\[ E_{\text{net}} = (18 \times 10^{3}) + (18 \times 10^{3}) = 36 \times 10^{3} \text{ N/C} \]
The field does not cancel out because the charges are of opposite signs.
Step 4: Final Answer:
The net electric field at the midpoint is \( 36 \times 10^{3} \) N/C directed towards the negative charge.