Chapter 17, Problem 26P

To determine

A formula for $V_{BA}=V_B-V_A$ for point $B$ and $A$ on the line between the charges.

Answer to Problem 26P

The answer is $V_B-V_A=2kq\left(\frac{1}{d-b}-\frac{1}{b}\right)$ .

Explanation of Solution

Given:

Two equal bus opposite charges are separated by a distance d.

Formula used:

It is known that $V=k\frac{Q}{r}=\frac{1}{4\pi {\in }_0}\frac{Q}{r}$

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

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

Calculation:

The potential is a function of position, and it’s a sum of potentials of point charges at a given point.

Thus,

  $V_A=k\left(\frac{q}{b}+\frac{-q}{d-b}\right)$

  $V_B=k\left(\frac{q}{d-b}+\frac{-q}{b}\right)$

  $\Rightarrow V_B-V_A=2kq\left(\frac{1}{d-b}-\frac{1}{b}\right)$

Conclusion:

The required formula is $V_B-V_A=2kq\left(\frac{1}{d-b}-\frac{1}{b}\right)$ .