Derive an expression for electric potential at a point due to a system of $\mathrm{N}$ charges.
Derive an expression for electric potential at a point due to a system of $\mathrm{N}$ charges.
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 $i=1,2,3, \ldots, \mathrm{N}$
Similar Questions
The ratio of electric potentials at the point $E$ to that at the point $F$ is
The ratio of electric potentials at the point $E$ to that at the point $F$ is
Four electric charges $+q,+q, -q$ and $-q$ are placed at the comers of a square of side $2L$ (see figure). The electric potential at point $A,$ midway between the two charges $+q$ and $+q,$ is
Four electric charges $+q,+q, -q$ and $-q$ are placed at the comers of a square of side $2L$ (see figure). The electric potential at point $A,$ midway between the two charges $+q$ and $+q,$ is
- [AIPMT 2011]
Assertion: Electron move away from a region of higher potential to a region of lower potential.
Reason: An electron has a negative charge.
Assertion: Electron move away from a region of higher potential to a region of lower potential.
Reason: An electron has a negative charge.
- [AIIMS 1999]
The election field in a region is given by $\vec E = (Ax + B)\hat i$ where $E$ is in $N\,C^{-1}$ and $x$ in meters. The values of constants are $A = 20\, SI\, unit$ and $B = 10\, SI\, unit$. If the potential at $x =1$ is $V_1$ and that at $x = -5$ is $V_2$ then $V_1 -V_2$ is.....$V$
The election field in a region is given by $\vec E = (Ax + B)\hat i$ where $E$ is in $N\,C^{-1}$ and $x$ in meters. The values of constants are $A = 20\, SI\, unit$ and $B = 10\, SI\, unit$. If the potential at $x =1$ is $V_1$ and that at $x = -5$ is $V_2$ then $V_1 -V_2$ is.....$V$
- [JEE MAIN 2019]
Consider two charged metallic spheres $S_{1}$ and $\mathrm{S}_{2}$ of radii $\mathrm{R}_{1}$ and $\mathrm{R}_{2},$ respectively. The electric $\left.\text { fields }\left.\mathrm{E}_{1} \text { (on } \mathrm{S}_{1}\right) \text { and } \mathrm{E}_{2} \text { (on } \mathrm{S}_{2}\right)$ on their surfaces are such that $\mathrm{E}_{1} / \mathrm{E}_{2}=\mathrm{R}_{1} / \mathrm{R}_{2} .$ Then the ratio $\left.\mathrm{V}_{1}\left(\mathrm{on}\; \mathrm{S}_{1}\right) / \mathrm{V}_{2} \text { (on } \mathrm{S}_{2}\right)$ of the electrostatic potentials on each sphere is
Consider two charged metallic spheres $S_{1}$ and $\mathrm{S}_{2}$ of radii $\mathrm{R}_{1}$ and $\mathrm{R}_{2},$ respectively. The electric $\left.\text { fields }\left.\mathrm{E}_{1} \text { (on } \mathrm{S}_{1}\right) \text { and } \mathrm{E}_{2} \text { (on } \mathrm{S}_{2}\right)$ on their surfaces are such that $\mathrm{E}_{1} / \mathrm{E}_{2}=\mathrm{R}_{1} / \mathrm{R}_{2} .$ Then the ratio $\left.\mathrm{V}_{1}\left(\mathrm{on}\; \mathrm{S}_{1}\right) / \mathrm{V}_{2} \text { (on } \mathrm{S}_{2}\right)$ of the electrostatic potentials on each sphere is
- [JEE MAIN 2020]