Chapter 17, Problem 19P

To determine

The potential at the fourth corner (point A).

Answer to Problem 19P

  $\frac{\sqrt{2}kQ}{2L}\left(\sqrt{2}+1\right)$

Explanation of Solution

Given:

The figure:

Potential, $V=0$

Formula used:

It is known that

  $V=k\frac{Q}{r}$

or

Expression for the Potential $V$ due to a charge $Q$ at a distance $L$ : $V=k\frac{Q}{L}$

Where, electrostatic constant, $k=8.99\times {10}^{9}N\cdot m^2/C^2$

  $V$ is absolute potential at a distance $r$ from the charge $Q$ .

Calculation:

Naming the points in a counter-clockwise manner and calculating potential due to each charge:

  $V_B=-2k\frac{Q}{L}$

  $V_C=k\frac{Q}{\sqrt{2}L}$

  $V_D=3k\frac{Q}{L}$

The potential at a point $A$ is sum of individual potentials at that point due to three charges.

It known that potential is a scalar quantity then potentials will be algebraically summed. So,

  $\begin{array}{l}{V}_{total}=V_B+V_C+V_D\\ =\left(-2k\frac{Q}{L}\right)+\left(k\frac{Q}{\sqrt{2}L}\right)+\left(3k\frac{Q}{L}\right)\end{array}$

According to the question, at very large distance, $V=0$

As the distance will be increase and tend to infinity, potential tend to zero. That means, ${\lim }_{r\to \infty }k\frac{Q}{r}=0$ . So,

  $V_{total}=\frac{\sqrt{2}kQ}{2L}\left(\sqrt{2}+1\right)$

Conclusion:

Potential at point $A$ is the sum of potentials due to each of the charges.