Chapter 8, Problem 8.1P

A point charge, Q = - 0.3 /C and m = 3 Ă— -10^-16 kg, is moving through the field E = 3₀a_z V/m. Use Eq. (1) and Newton’s laws to develop the appropriate differential equations and solve them, subject to the initial conditions at t = 0,: v = 3 Ă— 10⁵ a_x m/s at the origin. At t = 3 μs, find (a) the position P(x,y, z) of the charge; (b) the velocity v; (c) the kinetic energy of the charge.

To determine

(a)

The position of charge at $t=3\mu \text{s}$.

Answer to Problem 8.1P

The position of charge at $t=3\mu \text{s}$ is $P\left(0.90,0,-0.135\right)$.

Explanation of Solution

Given Information:

The point charge having $Q=-0.3\mu \text{C}$ and $m=3\times {10}^{-16}\text{kg}$ is moving through the electric field $\text{E}=30{\text{a}}_z\text{V/m}$ . The point charge is at origin and $\text{v}=3\times {10}^5{\text{a}}_x\text{m/s}$ at $t=0$.

Calculation:

The force of the given point charge,

   $\begin{array}{l}\text{F}=Q\text{E}=-0.3\times {10}^{-6}\times 30{\text{a}}_z\\ =-9\times {10}^{-6}\text{N}\end{array}$

According to newtons law, this force is equal to,

   $\begin{array}{l}\text{F}=m\text{a}=-9\times {10}^{-6}\\ \Rightarrow 3\times {10}^{-16}\times \frac{d^2\text{z}}{d{t}^2}=-9\times {10}^{-6}\because \text{a}=\frac{d^2\text{z}}{d{t}^2}\\ \Rightarrow \frac{d^2\text{z}}{d{t}^2}=-3\times {10}^{10}\end{array}$

Now we can solve this differential equation,

   $\frac{d\text{z}}{dt}=-3\times {10}^{10}t+C_1$

At initial condition, the velocity is along x direction. So, $v_{\text{z}}=\frac{d\text{z}}{dt}=0$ at $t=0$ . So,

   $\begin{array}{l}{C}_1=0\Rightarrow \\ \frac{d\text{z}}{dt}=-3\times {10}^{10}t\\ \Rightarrow z=\frac{-3\times {10}^{10}{t}^2}{2}+C_2\end{array}$

But the initial position of charge is at origin. So,

   $\begin{array}{l}{C}_2=0\Rightarrow \\ z=\frac{-3\times {10}^{10}{t}^2}{2}=-1.5\times {10}^{10}{t}^2\text{m}\\ \Rightarrow z_{t=3\mu \text{s}}=-1.5\times {10}^{10}{\left(3\times {10}^{-6}\right)}^2\\ =-0.135\text{m}\end{array}$

The velocity along x axis,

   $\begin{array}{l}{v}_x=3\times {10}^5\text{m/s}\Rightarrow \\ x_{t=3\mu \text{s}}=3\times {10}^5\times 3\times {10}^{-6}\\ =0.90\text{m}\end{array}$

The velocity along y axis is 0. So, the y coordinate of new position remains 0.

Thus, the position of charge at $t=3\mu \text{s}$ ,

   $P\left(x,y,z\right)=P\left(0.90,0,-0.135\right)$.

Conclusion:

The position of charge at $t=3\mu \text{s}$ is $P\left(0.90,0,-0.135\right)$.

To determine

(b)

The velocity of charge at $t=3\mu \text{s}$.

Answer to Problem 8.1P

The velocity of charge at $t=3\mu \text{s}$ is, $\text{v}=3\times {10}^5{\text{a}}_x-9\times {10}^4{\text{a}}_z\text{m/s}$.

Explanation of Solution

Given Information:

The point charge having $Q=-0.3\mu \text{C}$ and $m=3\times {10}^{-16}\text{kg}$ is moving through the electric field $\text{E}=30{\text{a}}_z\text{V/m}$ . The point charge is at origin and $\text{v}=3\times {10}^5{\text{a}}_x\text{m/s}$ at $t=0$.

   $\frac{d\text{z}}{dt}=-3\times {10}^{10}t$.

Calculation:

The velocity component along z direction,

   $\begin{array}{l}{v}_z=\frac{d\text{z}}{dt}=-3\times {10}^{10}t\\ =-3\times {10}^{10}\times 3\times {10}^{-6}\\ =-9\times {10}^4\text{m/s}\end{array}$

The velocity component along x direction,

   ${\text{v}}_x=3\times {10}^5{\text{a}}_x\text{m/s}$

The velocity component along y direction is zero. So, the velocity vector,

   $\begin{array}{l}\text{v}=v_x{\text{a}}_x+v_y{\text{a}}_y+v_z{\text{a}}_z\\ =3\times {10}^5{\text{a}}_x+0{\text{a}}_y-9\times {10}^4{\text{a}}_z\\ =3\times {10}^5{\text{a}}_x-9\times {10}^4{\text{a}}_z\text{m/s}\end{array}$

Conclusion:

The velocity of charge at $t=3\mu \text{s}$ is, $\text{v}=3\times {10}^5{\text{a}}_x-9\times {10}^4{\text{a}}_z\text{m/s}$.

To determine

(c)

The kinetic energy of charge at $t=3\mu \text{s}$.

Answer to Problem 8.1P

The kinetic energy of charge at $t=3\mu \text{s}$ is, $1.5\times {10}^{-5}\text{J}$.

Explanation of Solution

Given Information:

The point charge having $Q=-0.3\mu \text{C}$ and $m=3\times {10}^{-16}\text{kg}$ is moving through the electric field $\text{E}=30{\text{a}}_z\text{V/m}$ . The point charge is at origin and $\text{v}=3\times {10}^5{\text{a}}_x\text{m/s}$ at $t=0$.

The velocity of charge at $t=3\mu \text{s}$ is, $\text{v}=3\times {10}^5{\text{a}}_x-9\times {10}^4{\text{a}}_z\text{m/s}$.

Calculation:

The kinetic enrgy,

   $\begin{array}{l}E=\frac{1}{2}m{v}^2=\frac{1}{2}\times 3\times {10}^{-16}\times {\left(\sqrt{{\left(3\times {10}^5\right)}^2+{\left(-9\times {10}^4\right)}^2}\right)}^2\\ =1.5\times {10}^{-5}\text{J}\end{array}$

Conclusion:

The kinetic energy of charge at $t=3\mu \text{s}$ is, $1.5\times {10}^{-5}\text{J}$.