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41P

Expert-verifiedFound in: Page 713

Book edition
10th Edition

Author(s)
David Halliday

Pages
1328 pages

ISBN
9781118230718

**A particle of charge ${\mathbf{+}}{\mathbf{7}}{\mathbf{.}}{\mathbf{5}}{\mathbf{}}{\mathbf{\mu C}}$**** is released from rest at the point ${\mathbf{x}}{\mathbf{=}}{\mathbf{60}}{\mathbf{}}{\mathbf{cm}}$**** on an x-axis. The particle begins to move due to the presence of a charge Q that remains fixed at the origin. What is the kinetic energy of the particle at the instant it has moved 40 cm **

- The kinetic energy of the particle at the instant if $\mathrm{Q}=+20\mathrm{\mu C}$ is $0.90\mathrm{J}$.
- The kinetic energy of the particle at the instant if $\mathrm{Q}=-20\mathrm{\mu C}$ is $4.5\mathrm{J}$.

- Charge of the particle, $\mathrm{q}=+7.5\times {10}^{-6}\mathrm{C}$
- The particle is released from rest at the point $\mathrm{r}=0.6\mathrm{m}$ on the x-axis
- The distance at which it is moved,

**Using the concept of the conservation of energy and the formula of the electric potential energy, we can get the value of the final kinetic energy of the particle at the instant for different values of the charges.**

Formulae:

The potential energy of the system due to point charges, $\mathrm{U}=\frac{\mathrm{qQ}}{4{\mathrm{\pi \epsilon}}_{\mathrm{o}}\mathrm{r}}$ (i)

Applying to the law of conservation of energy, ${\mathrm{U}}_{\mathrm{o}}+{\mathrm{K}}_{\mathrm{o}}={\mathrm{U}}_{\mathrm{f}}+{\mathrm{K}}_{\mathrm{f}}$ (ii)

We have to apply conservation of energy to the particle with charge, which has zero initial kinetic energy. Thus, using this data in equation (ii), we get the equation of the final kinetic energy as follows:

${\mathrm{U}}_{\mathrm{o}}={\mathrm{U}}_{\mathrm{f}}+{\mathrm{K}}_{\mathrm{f}}\phantom{\rule{0ex}{0ex}}{\mathrm{K}}_{\mathrm{f}}={\mathrm{U}}_{\mathrm{o}}-{\mathrm{U}}_{\mathrm{f}}................\left(\mathrm{a}\right)$

The initial total energy of the particle is given using equation (i) as follows:

$\begin{array}{rcl}{\mathrm{U}}_{\mathrm{o}}& =& \frac{\left(9\times {10}^{9}\right)\left(+7.5\times {10}^{-6}\mathrm{C}\right)\left(+20\times {10}^{-6}\mathrm{C}\right)}{0.60\mathrm{m}}\\ & =& 2.25\mathrm{J}\end{array}$

Since the particles repel each other the final separation distance between them is given as:

$\left(0.60\mathrm{m}+0.40\mathrm{m}\right)=1.0\mathrm{m}$

Potential energy at final position is given using equation (i) as follows:

$\begin{array}{rcl}{\mathrm{U}}_{\mathrm{f}}& =& \frac{\left(9\times {10}^{9}\right)\left(+7.5\times {10}^{-6}\mathrm{C}\right)\left(+20\times {10}^{-6}\mathrm{C}\right)}{1.0\mathrm{m}}\\ & =& 1.35\mathrm{J}\end{array}$

Thus, the required kinetic energy at final position is given using equation (a) as follows:

role="math" localid="1662608251394" $\begin{array}{rcl}{\mathrm{K}}_{\mathrm{f}}& =& \left(2.25-1.35\right)\\ & =& 0.90\mathrm{J}\end{array}$

Hence, the value of the kinetic energy is $0.90\mathrm{J}$.

If the charge of the particle is $\mathrm{Q}=-20\mathrm{\mu C}$

Now the particles attract each other so the final separation between them is:

$\begin{array}{rcl}{\mathrm{r}}_{\mathrm{f}}& =& 0.60\mathrm{m}-0.40\mathrm{m}\\ & =& 0.20\mathrm{m}\end{array}$

Potential energy at the final position is given using equation (i) as follows:

$\begin{array}{rcl}{\mathrm{U}}_{\mathrm{f}}& =& \frac{\left(9\times {10}^{9}\right)\left(+7.5\times {10}^{-6}\mathrm{C}\right)\left(-20\times {10}^{-6}\mathrm{C}\right)}{0.20\mathrm{m}}\mathrm{J}\\ & =& -6.75\mathrm{J}\end{array}$

Now, using the data in equation (a), the required kinetic energy at the instant is given as:

$\begin{array}{rcl}{\mathrm{K}}_{\mathrm{f}}& =& \left(-2.25\mathrm{J}\right)-\left(-6.75\mathrm{J}\right)\\ & =& 4.5\mathrm{J}\end{array}$

Hence, the value of the energy is 4.5 J.

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