Question

Figure 24-29 shows four arrangements of charged particles, all the same distance from the origin. Rank the situations according to the net electric potential at the origin, most positive first. Take the potential to be zero at infinity.

Short answer

The rank of the situations according to the net electric potential at the origin is V₂ > V₁ >= V₃ = V₄.

Step-by-step solution

The given data

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

Understanding the concept of electric potential

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,

$\mathit{V}\mathbf{=}\frac{\mathbf{k}\mathbf{q}}{\mathbf{r}}$ ………………..(i)

Calculation of the rank of the situations according to the net electric potential

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.