# 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

$$