# ELECTRIC POTENTIAL DUE TO SYSTEM OF CHARGES

In this article we are going to derive an expression for electric potential due to system of charges. So keep reading…

### DERIVATION FOR THE EXPRESSION [latexpage]
Let’s consider $\displaystyle{q_1, q_2, q_3, … q_n}$ are the point charges which are located at the distances $\displaystyle{r_1, r_2, r_3 ,… r_n}$ respectively from a point P. See figure above. Now electric potential due to charge $q_1$ is-
$$V_1=\frac{1}{4\pi\epsilon_{0}}\frac{q_1}{r_1}$$
Similarly, electric potential due to others charges.
\begin{align*}
V_{2}&=\frac{1}{4\pi\epsilon_{0}}\frac{q_2}{r_2}\\
V_{3}&=\frac{1}{4\pi\epsilon_{0}}\frac{q_3}{r_3}\\
V_{n}&=\frac{1}{4\pi\epsilon_{0}}\frac{q_n}{r_n}
\end{align*}
By using superposition principle, we can obtain resultant value of electric potential at point P due to total charge configuration as algebraic sum of all the electric potential due to individuals charges.
\begin{align*}
V&=V_1+V_2+V_3+…+V_n\\
V&=\frac{1}{4\pi\epsilon_{0}}\frac{q_1}{r_1}+\frac{1}{4\pi\epsilon_{0}}\frac{q_2}{r_2}\\
&+\frac{1}{4\pi\epsilon_{0}}\frac{q_3}{r_3}+…+\frac{1}{4\pi\epsilon_{0}}\frac{q_n}{r_n}
\end{align*}
Now taking common $\displaystyle{\frac{1}{4\pi\epsilon_{0}}}$ from above equation, then we get-
$$V=\frac{1}{4\pi\epsilon_{0}}\left[\frac{q_1}{r_1}+\frac{q_2}{r_2}+\frac{q_3}{r_3}+…+\frac{q_n}{r_n}\right]$$
$$\implies \boxed{V=\frac{1}{4\pi\epsilon_{0}}\sum_{i=1}^{n}\frac{q_i}{r_i}}$$
The net electric potential due to the multiple charges at the point P is equal to the algebraic sum of all the potential due to individuals charges. Mathematically it is expressed as-
$$V_{net}=\sum_{i=1}^{n}V_i$$