2. Electric Potential and Capacitance
medium

Derive an expression for electric potential at a point due to a system of $\mathrm{N}$ charges.

Option A
Option B
Option C
Option D

Solution

Consider a system of charges $q_{1}, q_{2}, q_{3}, \ldots, q_{\mathrm{N}}$ with position vectors $r_{1}, r_{2}, r_{3}, \ldots, r_{\mathrm{N}}$

Electric potential at point $\mathrm{P}$ due to charge $q_{1}$,

$\mathrm{V}_{1}=\frac{k q_{1}}{r_{1 p}}$

where $k$ is coulomb constant $=\frac{1}{4 \pi \epsilon_{0}}$ and

$r_{1} \mathrm{P}=$ distance between charge $q_{1}$ and point $\mathrm{P}$.

Similarly electric potential due to charges $q_{2}, q_{3}, \ldots, q_{\mathrm{N}}$ are

$\mathrm{V}_{2}=\frac{k q_{2}}{r_{2 \mathrm{p}}}, \mathrm{V}_{3}=\frac{k q_{3}}{r_{3 \mathrm{p}}}$ and $\mathrm{V}_{\mathrm{N}}=\frac{k q_{\mathrm{N}}}{r_{\mathrm{NP}}}$

Electric potential is a scalar quantity. Hence total electric potential at $\mathrm{P}$ is, $\mathrm{V}=\mathrm{V}_{1}+\mathrm{V}_{2}+\mathrm{V}_{3}+\ldots ., \mathrm{V}_{\mathrm{N}}$

$\therefore \mathrm{V}=k\left[\frac{q_{1}}{r_{\mathrm{IP}}}+\frac{q_{2}}{r_{2 \mathrm{P}}}+\frac{q_{3}}{r_{3 \mathrm{P}}}+\frac{q_{\mathrm{N}}}{r_{\mathrm{NP}}}\right]$

$\therefore \mathrm{V}=k \sum_{i=1}^{\mathrm{N}} \frac{q_{i}}{r_{i \mathrm{P}}} \quad$ where $i=1,2,3, \ldots, \mathrm{N}$

Standard 12
Physics

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