Chapter 17, Problem 36QAP

To determine

(a)

The potential energy of the given system of charges.

Answer to Problem 36QAP

The potential energy of the given system of charges is $5.625\text{mJ}$.

Explanation of Solution

Given info:

Charge $q_0=0.50\mu \text{C}$

Charge $q=1.0\mu \text{C}$

Position of charge $q_0$is, $x_0=0$

Position of charge $q$is, $x_1=0.80\text{m}$.

Formula used:

Formula to calculate the potential energy of two charges is given as,

  $U=k\frac{q_{0}q}{\left(x_1-x\right)}$

Calculation:

Substituting the given values in the above equation, we get

  $\begin{array}{c}U=9\times {10}^9\times \frac{\left(0.50\times {10}^{-6}\right)\left(1.0\times {10}^{-6}\right)}{\left(0.8-0\right)}\\ U=5.625\times {10}^{-3}\text{J}\\ U=5.625\text{mJ}\end{array}$

Conclusion:

Thus, the potential energy of the given system of charges is $5.625\text{mJ}$.

To determine

(b)

The speed of particle with charge $q$when it reaches at $x=2.0\text{m}$.

Answer to Problem 36QAP

The speed of particle with charge $q$when it reaches at $x=2.0\text{m}$is $14.5\text{m/s}$.

Explanation of Solution

Given info:

Initial position of charge $q$is, $x_1=0.80\text{m}$.

Final position of charge $q$is, $x_2=2.0\text{m}$.

Initial velocity of charge $q$is, $u=0\text{m/s}$.

Mass of charge $q$is, $m=0.08\text{g}$.

Formula used:

Formula for the force experienced by the charged particle is,

  $F=k\frac{q_{0}q}{{\left(x_1-x\right)}^2}$

Formula for the third equation of motion is,

  $v^2=u^2+2a\Delta x$

Calculation:

The repulsive force experienced by particle with charge $q$can be calculated as,

  $\begin{array}{c}F=k\frac{q_{0}q}{{\left(x_1-x\right)}^2}\\ F=9\times {10}^9\times \frac{\left(0.50\times {10}^{-6}\right)\left(1.0\times {10}^{-6}\right)}{{\left(0.8-0\right)}^2}\\ F=7.03\times {10}^{-3}\text{N}\end{array}$

The acceleration of particle with charge $q$can be calculated as,

  $a=\frac{F}{m}$

The velocity of particle with charge $q$can be calculated as,

  $\begin{array}{c}{v}^2=u^2+2a\left(x_2-x_1\right)\\ v^2=u^2+2\left(\frac{F}{m}\right)\left(x_2-x_1\right)\end{array}$

Substituting the given values in the above equation, we get

  $\begin{array}{c}{v}^2=0+2\left(\frac{7.03\times {10}^{-3}}{0.08\times {10}^{-3}}\right)\left(2.0-0.80\right)\\ v^2=210.9\\ v=14.5\text{m/s}\end{array}$

Conclusion:

Thus, the speed of particle with charge $q$when it reaches at $x=2.0\text{m}$is $14.5\text{m/s}$.