Chapter 17, Problem 49P

To determine

Arrange the change in kinetic energy in the decreasing order.

Answer to Problem 49P

The decreasing order of change in kinetic energy is $\text{(c), (b), (e), (d), (a)}=\text{(f)}$.

Explanation of Solution

Write the expression for conservation of energy

$K_{\text{f}}+U_{\text{f}}=K_{\text{i}}+U_{\text{i}}$                                                              (I)

Here, $K_{\text{f}}$ is the final kinetic energy, $K_{\text{i}}$ is the initial kinetic energy, $U_{\text{f}}$ is the final potential energy and $U_{\text{i}}$ is the initial kinetic energy.

Rearrange (I)

$K_{\text{f}}-K_{\text{i}}=U_{\text{i}}-U_{\text{f}}$

Write the expression for change in kinetic energy

$\Delta K=-\Delta U$                                                  (II)

Here, $\Delta K$ is the change in kinetic energy and $\Delta U$ is the change in electric potential energy in moving the charge from $P$ to $S$.

Write the expression for $\Delta U$

$\Delta U=q\left(V_{\text{f}}-V_{\text{i}}\right)$                                                  (III)

Here, $q$ is the charge being moved, $V_{\text{f}}$ is the potential at $S$  and $V_{\text{i}}$ is the potential at the point, $P$.

Substitute (III) in (II)

$\Delta K=-q\left(V_{\text{f}}-V_{\text{i}}\right)$                                                  (IV)

(a)Substitute  $-5\text{nC}$ for $q$, $100\text{V}$ for $V_{\text{i}}$ and $-50\text{V}$ for $V_{\text{f}}$ in (IV) to find $\Delta K$

$\begin{array}{c}\Delta K=-\left(-5\text{nC}\right)\left(-50\text{V}-\left(100\text{V}\right)\right)\\ =-\left(-5\times {10}^{-9}\text{C}\right)\left(-50\text{V}-\left(100\text{V}\right)\right)\\ =-750\times {10}^{-9}\text{ J}\end{array}$

Thus, the change in kinetic energy in (a) is $-750\times {10}^{-9}\text{ J}$.

(b)Substitute  $-5\text{nC}$ for $q$, $-50\text{V}$ for $V_{\text{i}}$ and $50\text{V}$ for $V_{\text{f}}$ in (IV) to find $\Delta K$

$\begin{array}{c}\Delta K=-\left(-5\text{nC}\right)\left(50\text{V}-\left(-50\text{V}\right)\right)\\ =-\left(-5\times {10}^{-9}\text{C}\right)\left(50\text{V}-\left(-50\text{V}\right)\right)\\ =500\times {10}^{-9}\text{ J}\end{array}$

Thus, the change in kinetic energy in (b) is $500\times {10}^{-9}\text{ J}$.

(c)Substitute  $25\text{nC}$ for $q$, $50\text{V}$ for $V_{\text{i}}$ and $20\text{V}$ for $V_{\text{f}}$ in (IV) to find $\Delta K$

$\begin{array}{c}\Delta K=-\left(25\text{nC}\right)\left(20\text{V}-\left(50\text{V}\right)\right)\\ =-\left(25\times {10}^{-9}\text{C}\right)\left(20\text{V}-\left(50\text{V}\right)\right)\\ =750\times {10}^{-9}\text{ J}\end{array}$

Thus, the change in kinetic energy in (c) is $750\times {10}^{-9}\text{ J}$.

(d)Substitute  $-1\text{nC}$ for $q$, $400\text{V}$ for $V_{\text{i}}$ and $-100\text{V}$ for $V_{\text{f}}$ in (IV) to find $\Delta K$

$\begin{array}{c}\Delta K=-\left(-1\text{nC}\right)\left(-100\text{V}-\left(400\text{V}\right)\right)\\ =-\left(-1\times {10}^{-9}\text{C}\right)\left(-100\text{V}-\left(400\text{V}\right)\right)\\ =-500\times {10}^{-9}\text{ J}\end{array}$

Thus, the change in kinetic energy in (d) is $-500\times {10}^{-9}\text{ J}$.

(e)Substitute  $1\text{nC}$ for $q$, $-100\text{V}$ for $V_{\text{i}}$ and $-250\text{V}$ for $V_{\text{f}}$ in (IV) to find $\Delta K$

$\begin{array}{c}\Delta K=-\left(1\text{nC}\right)\left(-250\text{V}-\left(-100\text{V}\right)\right)\\ =-\left(1\times {10}^{-9}\text{C}\right)\left(-250\text{V}-\left(-100\text{V}\right)\right)\\ =150\times {10}^{-9}\text{ J}\end{array}$

Thus, the change in kinetic energy in (e) is $150\times {10}^{-9}\text{ J}$.

(e)Substitute  $5\text{nC}$ for $q$, $100\text{V}$ for $V_{\text{i}}$ and $250\text{V}$ for $V_{\text{f}}$ in (IV) to find $\Delta K$

$\begin{array}{c}\Delta K=-\left(5\text{nC}\right)\left(250\text{V}-\left(100\text{V}\right)\right)\\ =-\left(5\times {10}^{-9}\text{C}\right)\left(250\text{V}-\left(100\text{V}\right)\right)\\ =-750\times {10}^{-9}\text{ J}\end{array}$

Thus, the change in kinetic energy in (f) is $-750\times {10}^{-9}\text{ J}$.

Thus, the decreasing order of change in kinetic energy is $\text{(c), (b), (e), (d), (a)}=\text{(f)}$.