Question:
Two point charges, \(+2\,\mu\text{C}\) and \(-2\,\mu\text{C}\), are fixed respectively at points \(A\) and \(B\), separated by \(4\,\text{m}\) in air. A point \(P\) is located such that it is \(3\,\text{m}\) from \(+2\,\mu\text{C}\) and \(5\,\text{m}\) from \(-2\,\mu\text{C}\). What is the electric potential at point \(P\)?
\[
k=9\times10^9\,\text{N m}^2\text{C}^{-2}
\]
Two point charges, \(+2\,\mu\text{C}\) and \(-2\,\mu\text{C}\), are fixed respectively at points \(A\) and \(B\), separated by \(4\,\text{m}\) in air. A point \(P\) is located such that it is \(3\,\text{m}\) from \(+2\,\mu\text{C}\) and \(5\,\text{m}\) from \(-2\,\mu\text{C}\). What is the electric potential at point \(P\)?
\[
k=9\times10^9\,\text{N m}^2\text{C}^{-2}
\]
Show Hint
Electric potential is a scalar quantity, so potentials are added algebraically:
\[
V=\sum \frac{kq_i}{r_i}.
\]
Positive charges contribute positive potential and negative charges contribute negative potential.
Updated On: Jul 9, 2026
- \(1.2\times10^3\,\text{V}\)
- \(2.4\times10^3\,\text{V}\)
- \(-3.6\times10^3\,\text{V}\)
- \(3.6\times10^3\,\text{V}\) \bigskip
Show Solution
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The Correct Option is B
Solution and Explanation
Concept:
The electric potential at a point due to a system of point charges is the algebraic sum of the potentials due to individual charges.
\[
V=\sum \frac{kq}{r}.
\]
Step 1: Calculate the potential due to the charge \(+2\,\mu\text{C}\). \[ V_1 = \frac{k(2\times10^{-6})}{3}. \] \[ V_1 = \frac{9\times10^9\times2\times10^{-6}}{3}. \] \[ V_1 = 6\times10^3\,\text{V}. \]
Step 2: Calculate the potential due to the charge \(-2\,\mu\text{C}\). \[ V_2 = \frac{k(-2\times10^{-6})}{5}. \] \[ V_2 = -\frac{9\times10^9\times2\times10^{-6}}{5}. \] \[ V_2 = -3.6\times10^3\,\text{V}. \]
Step 3: Find the resultant potential at point \(P\). \[ V = V_1+V_2. \] \[ V = 6\times10^3-3.6\times10^3. \] \[ V = 2.4\times10^3\,\text{V}. \]
Step 4: Write the final answer. \[ \boxed{V=2.4\times10^3\,\text{V}} \] \[ \boxed{\text{Answer = (B)}} \]
Step 1: Calculate the potential due to the charge \(+2\,\mu\text{C}\). \[ V_1 = \frac{k(2\times10^{-6})}{3}. \] \[ V_1 = \frac{9\times10^9\times2\times10^{-6}}{3}. \] \[ V_1 = 6\times10^3\,\text{V}. \]
Step 2: Calculate the potential due to the charge \(-2\,\mu\text{C}\). \[ V_2 = \frac{k(-2\times10^{-6})}{5}. \] \[ V_2 = -\frac{9\times10^9\times2\times10^{-6}}{5}. \] \[ V_2 = -3.6\times10^3\,\text{V}. \]
Step 3: Find the resultant potential at point \(P\). \[ V = V_1+V_2. \] \[ V = 6\times10^3-3.6\times10^3. \] \[ V = 2.4\times10^3\,\text{V}. \]
Step 4: Write the final answer. \[ \boxed{V=2.4\times10^3\,\text{V}} \] \[ \boxed{\text{Answer = (B)}} \]
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