Question:
Two point charges $q_1=4\mu C$ and $q_2=-4\mu C$ are located 40 cm apart in free space. The electric field at the mid point of the line joining $q_1$ and $q_2$ is
Two point charges $q_1=4\mu C$ and $q_2=-4\mu C$ are located 40 cm apart in free space. The electric field at the mid point of the line joining $q_1$ and $q_2$ is
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For equal and opposite charges, the field at the exact midpoint is always twice the field of a single charge, pointing towards the negative charge. If the charges were identical, the field at the midpoint would be zero.
Updated On: Jun 24, 2026
- \(9 \times 10^5 \text{NC}^{-1}\) towards $q_1$
- \(9 \times 10^5 \text{NC}^{-1}\) towards $q_2$
- \(18 \times 10^5 \text{NC}^{-1}\) towards $q_1$
- \(18 \times 10^5 \text{NC}^{-1}\) towards $q_2$
- zero
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The Correct Option is D
Solution and Explanation
Step 1: Understanding the Concept:
The electric field at a point is the vector sum of individual electric fields produced by each charge.
- Field from a positive charge is directed away from it.
- Field from a negative charge is directed towards it.
Step 2: Key Formula or Approach:
1. Magnitude \(E = k \frac{q}{r^2}\), where \(k = 9 \times 10^9 \text{ Nm}^2/\text{C}^2\).
2. Distance from charges to midpoint \(r = 40/2 = 20 \text{ cm} = 0.2 \text{ m}\).
Step 3: Detailed Explanation:
1. Field due to $q_1$ (positive):
\(E_1\) is directed away from $q_1$ (i.e., towards $q_2$).
\[ E_1 = \frac{9 \times 10^9 \cdot 4 \times 10^{-6}}{(0.2)^2} = \frac{36 \times 10^3}{0.04} = 9 \times 10^5 \text{ NC}^{-1} \]
2. Field due to $q_2$ (negative):
\(E_2\) is directed towards $q_2$.
Since magnitudes and distances are the same:
\[ E_2 = 9 \times 10^5 \text{ NC}^{-1} \]
3. Total Electric Field:
Both fields point in the same direction (towards $q_2$).
\[ E_{net} = E_1 + E_2 = 9 \times 10^5 + 9 \times 10^5 = 18 \times 10^5 \text{ NC}^{-1} \text{ towards } q_2 \]
Step 4: Final Answer:
The electric field is \(18 \times 10^5 \text{ NC}^{-1}\) towards $q_2$.
The electric field at a point is the vector sum of individual electric fields produced by each charge.
- Field from a positive charge is directed away from it.
- Field from a negative charge is directed towards it.
Step 2: Key Formula or Approach:
1. Magnitude \(E = k \frac{q}{r^2}\), where \(k = 9 \times 10^9 \text{ Nm}^2/\text{C}^2\).
2. Distance from charges to midpoint \(r = 40/2 = 20 \text{ cm} = 0.2 \text{ m}\).
Step 3: Detailed Explanation:
1. Field due to $q_1$ (positive):
\(E_1\) is directed away from $q_1$ (i.e., towards $q_2$).
\[ E_1 = \frac{9 \times 10^9 \cdot 4 \times 10^{-6}}{(0.2)^2} = \frac{36 \times 10^3}{0.04} = 9 \times 10^5 \text{ NC}^{-1} \]
2. Field due to $q_2$ (negative):
\(E_2\) is directed towards $q_2$.
Since magnitudes and distances are the same:
\[ E_2 = 9 \times 10^5 \text{ NC}^{-1} \]
3. Total Electric Field:
Both fields point in the same direction (towards $q_2$).
\[ E_{net} = E_1 + E_2 = 9 \times 10^5 + 9 \times 10^5 = 18 \times 10^5 \text{ NC}^{-1} \text{ towards } q_2 \]
Step 4: Final Answer:
The electric field is \(18 \times 10^5 \text{ NC}^{-1}\) towards $q_2$.
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