91Ó°ÊÓ

Figure 24-29 gives four arrangements of charged particles, all at the same distance from origin.

Here, the net electric potential is given by the sum of all the individual potentials that are given by the distance of each point from the center of the square.

The electric potential at a point due to an individual charge,

$\mathit{V}\mathbf{=}\frac{\mathbf{k}\mathbf{q}}{\mathbf{r}}$ …â¶Ä¦â¶Ä¦â¶Ä¦â¶Ä¦â¶Ä¦..(¾±)

Let the distance of all the particles from origin be r.

Thus, the net potential at origin for all the four arrangements can be given using equation (i) as follows:

For arrangement-1:

$\begin{array}{rcl}{V}_1& =& \frac{k\left(2q\right)}{r}+\frac{k\left(-9q\right)}{r}\\ & =& \frac{-k\left(7q\right)}{r}\\ V_2& =& \frac{k\left(-2q\right)}{r}+\frac{k\left(-3q\right)}{r}\\ & =& \frac{-k\left(5q\right)}{r}\\ V_3& =& \frac{k\left(-2q\right)}{r}+\frac{k\left(-2q\right)}{r}+\frac{k\left(-q\right)}{r}+\frac{k\left(-2q\right)}{r}\\ & =& \frac{-k\left(7q\right)}{r}\\ V_4& =& \frac{k\left(2q\right)}{r}+\frac{k\left(-4q\right)}{r}+\frac{k\left(2q\right)}{r}+\frac{k\left(-7q\right)}{r}\\ & =& \frac{-k\left(7q\right)}{r}\end{array}$

Hence, the rank of the arrangements according to their net potentials is V₂ > V₁ = V₃ = V₄.