Chapter 17, Problem 72GP

To determine

The change in potential energy.

Answer to Problem 72GP

  $\text{1}\text{.5}\text{J}$

Explanation of Solution

Given:

  $\text{33}\text{μC}$ point charge is placed $\text{36}\text{cm}$ from identical $\text{+33}\text{μC}$ charge.

  $\text{-1}\text{.5}\text{μC}$ charge is moved from A point to B point.

Formula used:

The potential energy is given by

  $\text{PE = qV}$

Where,

  $\text{q}$ is the charge of an electron.

  $\text{V}$ is the voltage applied.

The potential is a distance of $\text{r}$ from a point change $\text{Q}$ can be expressed as $\text{V}\text{=}\frac{\text{kQ}}{\text{r}}$

Here, $\text{k}$ is the Coulomb’s constant $\left(\text{9}\text{.0}\text{×}{\text{10}}^{\text{9}}\text{N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}\right)$

Calculation:

Suppose that ${\text{d}}_{\text{1}}$ represent the distance from the left charge to point $\text{b}$ , and let ${\text{d}}_{\text{2}}$ represent the distance from the right charge to point $\text{b}$ .

Assume $\text{Q}$ represents the positive charge, and $\text{q}$ represents the negative charge that moves.

So, the change in potential energy will be ${\text{PE}}_{\text{b}}{\text{-PE}}_{\text{a}}{\text{=qV}}_{\text{ba}}\text{=q}\left({\text{V}}_{\text{b}}{\text{-V}}_{\text{a}}\right)\mathrm{...}\left(1\right)$

Now, calculate the diameter:

  ${\text{d}}_{\text{1}}\text{=}\sqrt{{\text{12}}^{\text{2}}{\text{+14}}^{\text{2}}}\text{cm}\text{=}\text{18}\text{.44}\text{cm}$ and ${\text{d}}_{\text{2}}\text{=}\sqrt{{\text{14}}^{\text{2}}{\text{+24}}^{\text{2}}}\text{cm}\text{=}\text{27}\text{.78}\text{cm}$

Substitute the values in equation $\left(1\right)$

  ${\text{PE}}_{\text{b}}{\text{-PE}}_{\text{a}}\text{=q}\left({\text{V}}_{\text{b}}{\text{-V}}_{\text{a}}\right)\text{=q}\left[\left(\frac{\text{kQ}}{\text{0}\text{.1844}\text{m}}\text{+}\frac{\text{kQ}}{\text{0}\text{.2778}\text{m}}\right)\text{-}\left(\frac{\text{kQ}}{\text{0}\text{.12}\text{m}}\text{+}\frac{\text{kQ}}{\text{0}\text{.24}\text{m}}\right)\right]$

  $\text{=}\text{kQq}\left[\left(\frac{\text{1}}{\text{0}\text{.1844}\text{m}}\text{+}\frac{\text{1}}{\text{0}\text{.2778}\text{m}}\right)\text{-}\left(\frac{\text{1}}{\text{0}\text{.12}\text{m}}\text{+}\frac{\text{1}}{\text{0}\text{.24}\text{m}}\right)\right]$

  $\begin{array}{l}\text{=}\left(\text{8}\text{.99}\times {\text{10}}^{\text{9}}\right)\left(\text{-1}{\text{.5×10}}^{\text{-6}}\text{C}\right)\left({\text{33×10}}^{\text{-6}}\text{C}\right)\left(\text{-3}\text{.477}{\text{m}}^{\text{-1}}\right)\\ \text{=}\text{1}\text{.547}\text{J}\approx 1.5\text{J}\end{array}$

Conclusion:

The required value is $1.5\text{J}$ .