Question

Question: Three charges are at the corners of an equilateral triangle (side l) as shown in Fig. 17–45. Determine the potential at the midpoint of each of the sides. Let \[{\bf{V = 0}}\] at \[{\bf{r = }}\infty \].

FIGURE 17–45 Problem 75.

Short answer

The electric potentials at the middle of the sides a, b, and c are\[ - 6.85\frac{{kQ}}{l}\], \[ - 3.46\frac{{kQ}}{l}\], and \[ - 5.15\frac{{kQ}}{l}\] respectively.

Step-by-step solution

Step 1:Variables on which the electric potential depends 

The electric potential value relies on the magnitude of the charge and the distance of the charge from a specific point.

The electric potential is given by,

\(V = \frac{{kQ}}{r}\)

Here, k is the Coulomb’s constant, Q is the charge and r is the distance of the point from the charge

Evaluation of the electric potential at the middle of the side of an equilateral triangle

The schematic diagram for the problem can be drawn as:

The electric potential at the middle of the side a can be calculated as:

\[\begin{array}{c}{V_{\rm{a}}} = k\frac{{\left( { - Q} \right)}}{{\left( {\frac{l}{2}} \right)}} + k\frac{{\left( { - 3Q} \right)}}{{\left( {\frac{l}{2}} \right)}} + k\frac{{\left( Q \right)}}{{\left( {\frac{{\sqrt 3 l}}{2}} \right)}}\\ = \frac{{2kQ}}{l}\left( { - 1 - 3 + \frac{1}{{\sqrt 3 }}} \right)\\ = - 6.85\frac{{kQ}}{l}\end{array}\]

The electric potential at the middle of the side b can be calculated as:

\[\begin{array}{c}{V_{\rm{b}}} = k\frac{{\left( { - Q} \right)}}{{\left( {\frac{l}{2}} \right)}} + k\frac{{\left( Q \right)}}{{\left( {\frac{l}{2}} \right)}} + k\frac{{\left( { - 3Q} \right)}}{{\left( {\frac{{\sqrt 3 l}}{2}} \right)}}\\ = \frac{{2kQ}}{l}\left( { - 1 + 1 - \frac{3}{{\sqrt 3 }}} \right)\\ = - 3.46\frac{{kQ}}{l}\end{array}\]

The electric potential at the middle of the side c can be calculated as:

\[\begin{array}{c}{V_{\rm{c}}} = \frac{{k\left( Q \right)}}{{\left( {\frac{l}{2}} \right)}} + k\frac{{\left( { - 3Q} \right)}}{{\left( {\frac{l}{2}} \right)}} + \frac{{k\left( { - Q} \right)}}{{\left( {\frac{{\sqrt 3 l}}{2}} \right)}}\\ = \frac{{2kQ}}{l}\left[ {1 - 3 - \frac{1}{{\sqrt 3 }}} \right]\\ = - 5.15\frac{{kQ}}{l}\end{array}\]

Thus, the electric potentials at the middle of the sides a, b, and c are\[ - 6.85\frac{{kQ}}{l}\], \[ - 3.46\frac{{kQ}}{l}\], and \[ - 5.15\frac{{kQ}}{l}\] respectively.