Chapter 17, Problem 62GP

To determine

The potential at the midpoint of each side of the equilateral triangle.

Answer to Problem 62GP

Potential at point a is ${\text{V}}_{\text{a}}\text{=}\text{-6}\text{.85}\frac{\text{kQ}}{\text{L}}$

Potential at point b is ${\text{V}}_{\text{b}}\text{=}\text{-3}\text{.46}\frac{\text{kQ}}{\text{L}}$

Potential at point c is ${\text{V}}_{\text{c}}\text{=-5}\text{.15}\frac{\text{kQ}}{\text{L}}$

Explanation of Solution

Given:

There are three charges on the corners of the equilateral triangle.

Formula used:

The potential at a point is given by $\frac{\text{kQ}}{\text{r}}$

Where, $\text{k}$ is Coulomb constant $\text{k}\text{=}\frac{\text{1}}{{\text{4πε}}_{\text{0}}}\text{=}\text{constant}$

  $\text{Q}$ is the charge of an electron.

  $\text{r}$ is the distance.

Calculation:

According to the question, the triangle is equilateral. So, the distance from a corner to the midpoint of the opposite side is $\sqrt{\text{3}}\text{L/2}$

  

The potential at the point a is

  ${\text{V}}_{\text{a}}\text{=}\frac{\text{k}\left(\text{-Q}\right)}{\frac{\text{L}}{\text{2}}}\text{+}\frac{\text{k}\left(\text{-3Q}\right)}{\frac{\text{L}}{\text{2}}}\text{+}\frac{\text{k}\left(\text{-3Q}\right)}{\frac{\sqrt{\text{3}}\text{L}}{\text{2}}}$

  ${\text{V}}_{\text{a}}\text{=}\frac{\text{2kQ}}{\text{L}}\left[\text{-4+}\frac{1}{\sqrt{3}}\right]$

  ${\text{V}}_{\text{a}}\text{=}\text{-6}\text{.85}\frac{\text{kQ}}{\text{L}}$

The potential at the point b is

  ${\text{V}}_{\text{b}}\text{=}\frac{\text{k}\left(\text{-Q}\right)}{\frac{\text{L}}{\text{2}}}\text{+}\frac{\text{k}\left(\text{Q}\right)}{\frac{\text{L}}{\text{2}}}\text{+}\frac{\text{k}\left(\text{-3Q}\right)}{\frac{\sqrt{\text{3}}\text{L}}{\text{2}}}$

  ${\text{V}}_{\text{b}}\text{=}\frac{\text{6kQ}}{\sqrt{\text{3}}\text{L}}\text{=}\text{-3}\text{.46}\frac{\text{kQ}}{\text{L}}$

The potential at the point c is

  ${\text{V}}_{\text{c}}\text{=}\frac{\text{k}\left(\text{-Q}\right)}{\frac{\text{L}}{\text{2}}}\text{+}\frac{\text{k}\left(\text{-3Q}\right)}{\frac{\text{L}}{\text{2}}}\text{+}\frac{\text{k}\left(\text{-Q}\right)}{\frac{\sqrt{\text{3}}\text{L}}{\text{2}}}$

  ${\text{V}}_{\text{c}}\text{=-}\frac{\text{2kQ}}{\text{L}}\left[\text{2+}\frac{1}{\sqrt{3}}\right]$

  ${\text{V}}_{\text{c}}\text{=-5}\text{.15}\frac{\text{kQ}}{\text{L}}$

Conclusion:

All the required values are calculated.