Question:
The equipotential surface of a system of two point charge $5\,\mu C$ and $-5\,\mu C$ at points $A$ and $B$ separated by $80$ cm is a plane perpendicular to the line connecting $A$ and $B$ at
The equipotential surface of a system of two point charge $5\,\mu C$ and $-5\,\mu C$ at points $A$ and $B$ separated by $80$ cm is a plane perpendicular to the line connecting $A$ and $B$ at
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For equal and opposite charges:
- Zero potential lies at midpoint
- Always look for symmetry
Updated On: Apr 30, 2026
- $0.4$ m from $A$
- $0.6$ m from $A$
- $0.5$ m from $A$
- $0.6$ m from $B$
- $0.5$ m from $B$
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The Correct Option is A
Solution and Explanation
Concept:
Equipotential surface means net potential is zero:
\[
V = k\left(\frac{q_1}{r_1} + \frac{q_2}{r_2}\right) = 0
\]
Step 1: Substitute charges.
\[ \frac{5}{r_1} - \frac{5}{r_2} = 0 \]
Step 2: Simplify.
\[ \frac{1}{r_1} = \frac{1}{r_2} \Rightarrow r_1 = r_2 \]
Step 3: Interpret physically.
The point must be equidistant from both charges.
Step 4: Find midpoint.
Distance between charges: \[ 80\ \text{cm} = 0.8\ \text{m} \] Midpoint: \[ \frac{0.8}{2} = 0.4\ \text{m} \]
Step 1: Substitute charges.
\[ \frac{5}{r_1} - \frac{5}{r_2} = 0 \]
Step 2: Simplify.
\[ \frac{1}{r_1} = \frac{1}{r_2} \Rightarrow r_1 = r_2 \]
Step 3: Interpret physically.
The point must be equidistant from both charges.
Step 4: Find midpoint.
Distance between charges: \[ 80\ \text{cm} = 0.8\ \text{m} \] Midpoint: \[ \frac{0.8}{2} = 0.4\ \text{m} \]
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