Question

(III) Two equal but opposite charges are separated by a distance d, as shown in Fig. 17–41. Determine a formula for \({{\bf{V}}_{{\bf{BA}}}}{\bf{ = }}{{\bf{V}}_{\bf{B}}}{\bf{ - }}{{\bf{V}}_{\bf{A}}}\)for points B and A on the line between the charges situated as shown.

FIGURE 17-41 Problem 30

Short answer

The required formula for electric potentialis\({V_{{\rm{BA}}}} = 2kq\left[ {\frac{{2b - d}}{{\left( {d - b} \right)b}}} \right]\).

Step-by-step solution

Understanding the electric potential

The electric potential energy per unit charge at any point is termed as the electric potential. It relies on the charge and the distance of the point from the charge.

The expression for electric potential is given as:

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

Here, k is the Coulomb’s constant, Q is the charge and r is the distance.

Evaluation of the electric potential at point A and B

The relation of electric potential at Ais given by,

\({V_{\rm{A}}} = \frac{{kq}}{b} + \frac{{k\left( { - q} \right)}}{{d - b}}\)

Here, kis the electric constant and bis the distance from the end point.

The relation of electric potential at Bis given by,

\({V_{\rm{B}}} = \frac{{kq}}{{d - b}} + \frac{{k\left( { - q} \right)}}{b}\)

Evaluation of the electric potential difference between charges

The relation of electric potential difference is given by,

\({V_{{\rm{BA}}}} = {V_{\rm{B}}} - {V_{\rm{A}}}\)

On plugging the values in the above relation.

\(\begin{aligned}{V_{{\rm{BA}}}} &= \left( {\left( {\frac{{kq}}{{d - b}} + \frac{{k\left( { - q} \right)}}{b}} \right) - \left( {\frac{{kq}}{b} + \frac{{k\left( { - q} \right)}}{{d - b}}} \right)} \right)\\{V_{{\rm{BA}}}} &= kq\left( {\frac{1}{{d - b}} - \frac{1}{b} - \frac{1}{b} + \frac{1}{{d - b}}} \right)\\{V_{{\rm{BA}}}} &= 2kq\left( {\frac{1}{{d - b}} - \frac{1}{b}} \right)\\{V_{{\rm{BA}}}} &= 2kq\left( {\frac{{2b - d}}{{\left( {d - b} \right)b}}} \right)\end{aligned}\)

Thus, the potential difference between points A and B is \({V_{{\rm{BA}}}} = 2kq\left( {\frac{{2b - d}}{{\left( {d - b} \right)b}}} \right)\).