Question:
Five equal point charges with charge $Q = 10$ nC are located at $x = 2, 4, 5, 10$ and $20$ m. If $\varepsilon_0 = \left[\dfrac{10^{-9}}{36\pi}\right]$ F/m, then the potential at the origin ($x = 0$) is
Five equal point charges with charge $Q = 10$ nC are located at $x = 2, 4, 5, 10$ and $20$ m. If $\varepsilon_0 = \left[\dfrac{10^{-9}}{36\pi}\right]$ F/m, then the potential at the origin ($x = 0$) is
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Potential is scalar, so directly add contributions without considering direction.
Updated On: May 1, 2026
- $9.9$ V
- $11.1$ V
- $90$ V
- $99$ V
- $111$ V
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The Correct Option is D
Solution and Explanation
Concept:
Electric potential: \[ V = \frac{1}{4\pi \varepsilon_0} \sum \frac{Q}{r} \]
Step 1: Find constant.
\[ \frac{1}{4\pi \varepsilon_0} = 9 \times 10^9 \]
Step 2: Substitute values.
\[ Q = 10 \text{ nC} = 10 \times 10^{-9} \] \[ V = 9 \times 10^9 \times 10 \times 10^{-9} \left(\frac{1}{2} + \frac{1}{4} + \frac{1}{5} + \frac{1}{10} + \frac{1}{20}\right) \]
Step 3: Simplify sum.
\[ = 90 \left(0.5 + 0.25 + 0.2 + 0.1 + 0.05\right) = 90 \times 1.1 = 99 \]
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