91Ó°ÊÓ

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

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.