Moving Charges!

a) Use Newton’s Second Law to solve for \(a\), and substitute the appropriate quantities for each particle.

b) Use the kinematics equation to solve for the position assuming the acceleration and the time are known variables.

c) Use the kinematics equation to solve for the final speed in terms of the acceleration and the given time.

a) Newton’s Second Law states:

\begin{equation*}
\sum \vec{F}=m\vec{a},
\end{equation*}

[mepr-show rules=”4409″ unauth=”both”]

where the only force is the electrical force which can be written as \(F_E = qE\). Solving for \(a\), we get:

\begin{equation*}
\vec{a}=\frac{qE}{m}\,\hat{\textbf{i}}.
\end{equation*}

With the given values for the negatively charged particle, we get:

\begin{equation*}
\vec{a}_{\text{neg}}\approx- 1.045\times 10^{13}\,\text{m/s}^2\,\hat{\textbf{i}}.
\end{equation*}

Whereas for the positively charged particle:

\begin{equation*}
\vec{a}_{\text{pos}}\approx 8.144\times 10^{9}\,\text{m/s}^2\,\hat{\textbf{i}}.
\end{equation*}

b) Using the kinematics equation:

\begin{equation*}
x_f=x_i+v_{x,i}t+\frac{1}{2}a_xt^2,
\end{equation*}

where it can be reduced to:

\begin{equation*}
x_f=\frac{1}{2}a_xt^2.
\end{equation*}

For the negatively charged particle:

\begin{equation*}
x_f^{\text{neg}}\approx -131\,\text{m}.
\end{equation*}

For the positively charged particle:

\begin{equation*}
x_f^{\text{pos}}\approx 0.102\,\text{m}.
\end{equation*}

c) To find the speed of each particle at a time when \(t\) has elapsed, we must use the kinematics equation for uniform acceleration:

\begin{equation}
\vec{v}_f=\vec{v}_i+\vec{a}t,
\end{equation}

Plugging in the given values for the negatively charged particle, we get:

\begin{equation*}
v_{x,f}^{\text{neg}}\approx -5.225\times 10^{7}\,\text{m/s}.
\end{equation*}

Plugging in the given values for the positively charged particle:

\begin{equation*}
v_{x,f}^{\text{pos}}\approx 4.072\times10^{4}\,\text{m/s}.
\end{equation*}

For a more detailed explanation of any of these steps, click on “Detailed Solution”.

[/mepr-show]

a) To calculate the acceleration, we must identify all the relevant forces exerted on the particles and then use Newton’s second law to find the acceleration, so let’s draw the forces exerted on each particle (see figure 1).

[mepr-show rules=”4409″ unauth=”both”]

The electric field \(\vec{E}\) can be written as its magnitude multiplied by the unitary vector indicating its direction; explicitly,

\begin{equation}
\label{evec}
\vec{E}=E\,\hat{\textbf{i}}.
\end{equation}

We can then write Newton’s second law as

\begin{equation}
\label{newton}
\sum \vec{F}=m\vec{a},
\end{equation}

where \(m\) is the mass of the particle, \(\vec{a}\) is the acceleration vector, and \(\sum \vec{F}\) is the sum of all forces exerted on the particle. For both particles the only relevant force will be the electric force \(\vec{F}_E\), so equation \eqref{newton} becomes

\begin{equation}
\label{newton2}
\vec{F}_E=m\vec{a}.
\end{equation}

The electric force is given by the charge multiplied by the electric field; namely,

\begin{equation}
\label{eforce}
\vec{F}_E=q\vec{E}.
\end{equation}

Using the expression for the electric force in equation \eqref{newton2}, we obtain

\begin{equation}
q\vec{E}=m\vec{a},
\end{equation}

where we can solve for the acceleration as

\begin{equation}
\label{acel}
\vec{a}=\frac{q}{m}\vec{E}.
\end{equation}

Using the expression for the electric field given in equation \eqref{evec} in equation \eqref{acel}, we get

\begin{equation}
\vec{a}=\frac{qE}{m}\,\hat{\textbf{i}}.
\end{equation}

For the negatively charged particle \(q=-6.8\times 10^{-18}\,\text{C}\) and \(m=500m_e\), where \(m_e=9.11\times10^{-31}\,\text{kg}\); so, its acceleration is

\begin{equation}
\vec{a}_{\text{neg}}=\frac{(-6.8\times 10^{-18}\,\text{C})(700\,\text{N/C})}{500(9.11\times10^{-31}\,\text{kg})}\,\hat{\textbf{i}},
\end{equation}

\begin{equation}
\label{aneg}
\vec{a}_{\text{neg}}\approx- 1.045\times 10^{13}\,\text{m/s}^2\,\hat{\textbf{i}}.
\end{equation}

Note here that the negative sign indicates that the direction of the negatively charged particle is towards the negative X-axis, meaning it is opposed to the electric field.

For the positively charge particle \(q=6.8\times 10^{-18}\,\text{C}\) and \(m=350m_p\), where \(m_p=1.67\times10^{-27}\,\text{kg}\); thus, its acceleration is

\begin{equation}
\vec{a}_{\text{pos}}=\frac{(6.8\times 10^{-18}\,\text{C})(700\,\text{N/C})}{350(1.67\times10^{-27}\,\text{kg})}\,\hat{\textbf{i}},
\end{equation}

\begin{equation}
\label{apos}
\vec{a}_{\text{pos}}\approx 8.144\times 10^{9}\,\text{m/s}^2\,\hat{\textbf{i}}.
\end{equation}

b) Now, let’s calculate the final position of each particle. From the results of acceleration found in part (a) and taking into account that the particles are released from rest we can state the following: the positive particle starts to move in the direction of the electric field \(\hat{\textbf{i}}\), while the negative charge starts to move in the direction opposite to the electric field, namely \(-\hat{\textbf{i}}\).

Suppose both particles start at \(x=0\). The position vector \(\vec{r}\) as a function of time \(t\) in the uniformly accelerated case can be written as

\begin{equation}
\vec{r}_f=\vec{r}_i+\vec{v}_it+\frac{1}{2}\vec{a}t^2,
\end{equation}

where \(\vec{v}_i\) is the initial velocity and \(\vec{r}_i\), \(\vec{r}_f\) are the initial and final position vectors respectively. Because the movement occurs only on the X axis, we can write the equation above as

\begin{equation}
x_f\,\hat{\textbf{i}}=x_i\,\hat{\textbf{i}}+v_{x,i}t\,\hat{\textbf{i}}+\frac{1}{2}a_x t^2\,\hat{\textbf{i}},
\end{equation}

or omitting the vector notation

\begin{equation}
\label{position}
x_f=x_i+v_{x,i}t+\frac{1}{2}a_xt^2,
\end{equation}

where \(a_x\) is the acceleration along the X axis. Because the initial velocity is zero and the initial position is also zero (we can assume that the particle is initially at the origin), we can write equation \eqref{position} as

\begin{equation}
\label{pos2}
x_f=\frac{1}{2}a_xt^2.
\end{equation}

For the negative particle we use the value for \(a_x=-1.045\times 10^{13}\,\text{m/s}^2\) as given in equation \eqref{aneg}. Equation \eqref{pos2} for the negative charge becomes

\begin{equation}
x_f^{\text{neg}}=\frac{1}{2}(-1.045\times 10^{13}\,\text{m/s}^2)(5\times 10^{-6}\,\text{s})^2,
\end{equation}

\begin{equation}
x_f^{\text{neg}}\approx -131\,\text{m}.
\end{equation}

For the positive charge, the acceleration is given by equation \eqref{apos}, so \(a_x=8.144\times 10^{9}\,\text{m/s}^2\); thus, using equation \eqref{pos2} we get

\begin{equation}
x_f^{\text{pos}}=\frac{1}{2}(8.144\times 10^{9}\,\text{m/s}^2)(5\times 10^{-6}\,\text{s})^2,
\end{equation}

\begin{equation}
x_f^{\text{pos}}\approx 0.102\,\text{m}.
\end{equation}

c) To find the speed of each particle when the time \(t\) has passed, we must use the cinematic relation for uniform acceleration

\begin{equation}
\vec{v}_f=\vec{v}_i+\vec{a}t,
\end{equation}

where \(\vec{v}_f\) is the final velocity and \(\vec{v}_i\) is the initial velocity. Because the movement is all along the X axis, we can write the equation above as

\begin{equation}
v_{x,f}\,\hat{\textbf{i}}=v_{x,i}\,\hat{\textbf{i}}+a_xt\,\hat{\textbf{i}}.
\end{equation}

Omitting the vector notation in the equation above, we obtain

\begin{equation}
v_{x,f}=v_{x,i}+a_xt,
\end{equation}

which can be simplified if we know that the initial velocity is zero; namely,

\begin{equation}
\label{velocity}
v_{x,f}=a_xt.
\end{equation}

Thus, for the negative charge, we have

\begin{equation}
v_{x,f}^{\text{neg}}=(-1.045\times 10^{13}\,\text{m/s}^2)(5\times 10^{-6}\,\text{s}),
\end{equation}

\begin{equation}
v_{x,f}^{\text{neg}}\approx -5.225\times 10^{7}\,\text{m/s}.
\end{equation}

Using equation \eqref{velocity} for the positive charge, we have

\begin{equation}
v_{x,f}^{\text{pos}}=(8.144\times 10^{9}\,\text{m/s}^2)(5\times 10^{-6}\,\text{s}),
\end{equation}

\begin{equation}
v_{x,f}^{\text{pos}}\approx 4.072\times10^{4}\,\text{m/s}.
\end{equation}

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