P19B2018 – Charged Rod

Note: to facilitate comprehension and avoid unnecessary confusion, we label $x_P$ the $x$-location of point $P$ instead of $x$.

The charge $Q$ is uniformly distributed over the length $L$ of the rod and we therefore introduce the linear charge density $\lambda $ which is equal to

\lambda =\frac{Q}{L}

We consider an infinitesimal slice $dx$ of the rod that carries an infinitesimal charge $dQ=\lambda dx$.

At point $P$, this charge $dQ$ (which is positive) creates an infinitesimal electric field $d{\overrightarrow{E}}_P$ along the $x$-axis (to the right) with magnitude $dE_P$ equal to

dE_P=\frac{kdQ}{{\left(x_P-x\right)}^2}=\frac{\lambda kdx}{{\left(x_P-x\right)}^2}=\frac{kQdx}{{\left(x_P-x\right)}^2L}

Since any arbitrary slice would create an electric field $d\overrightarrow{E}$ directed to the right at point $P$, we conclude that the electric field ${\overrightarrow{E}}_P$ at point $P$ is directed along the positive $x$-axis and has a magnitude $E_P$ given by integrating $dE$ over the rod as follows.

\boxed{E_P=\int^{x=0}_{x=-L}{\frac{kQdx}{{\left(x_P-x\right)}^2L}}={\left[\frac{kQ}{L\left(x_P-x\right)}\right]}^0_{-L}=\frac{kQ}{Lx_P}-\frac{kQ}{L\left(x_P+L\right)}=\frac{kQ}{x_P\left(x_P+L\right)}}

Note: the distance between the arbitrary slice and point $P$ is $x_P-x$ and not $x_P+x$ because $x$ is negative!

By Newton’s Third Law, the force exerted by the electron on the rod (hard to derive) is equal in magnitude to the force exerted by the rod on the electron (easy to derive).

The electron is negative and will therefore be attracted to the rod and the electric force $F_{rod\ on\ e}$ will point to the left. The magnitude of the force is equal to

F_{rod\ on\ e}=\left|-e\right|E_P=\frac{kQe}{x_P\left(x_P+L\right)}

Thus, the force exerted by the electron on the rod is directed to the right and has a magnitude equal to

\boxed{F_{e\ on\ rod}=\frac{kQe}{x_P\left(x_P+L\right)}}

The alpha particle would have to be located at a point on the $x$-axis such that the force it creates on the rod cancels with the force from the electron. This force would therefore have to point to the left (meaning the alpha particle would have to be to the right of the rod) and have a magnitude equal to $F_{e\ on\ rod}$.

Using Newton’s Third Law as we did in the previous question, we conclude that the rod would have exert a force on the alpha particle that is equal to $F_{e\ on\ rod}$ in magnitude (so that, in turn, the alpha particle could exert that amount of force on the rod).

If we let $x_a$ denote the location of the alpha particle, we know from question 1. that the electric field at that point has a magnitude $E_a$ equal to

E_a=\frac{kQ}{x_a\left(x_a+L\right)}

The force exerted by the rod on the alpha particle then has a magnitude $F_{rod\ on\ \alpha }$ equal to

F_{rod\ on\ \alpha }=2eE_a=\frac{2kQe}{x_a\left(x_a+L\right)}

Thus, we conclude that the electron exerts a force $F_{e\ on\ rod}$ on the rod that is directed to the right (and has the same magnitude as $F_{rod\ on\ \alpha }$) and the alpha particle exerts a force $F_{\alpha \ on\ rod}$ that is directed to the left. The net force on the rod will be zero if these two forces cancel and we therefore solve for $x_a$ as follows

\begin{aligned}
F_{e\ on\ rod}-F_{\alpha \ on\ rod}=0\ \ \ \ &\Rightarrow \ \ \ \ \ \frac{kQe}{x_P\left(x_P+L\right)}-\frac{2kQe}{x_a\left(x_a+L\right)}=0 \\
\\
&\Rightarrow \ \ \ \ \ \frac{1}{x_P\left(x_P+L\right)}=\frac{2}{x_a\left(x_a+L\right)} \\
\\
&\Rightarrow \ \ \ \ \ x_a\left(x_a+L\right)=2x_P\left(x_P+L\right) \\
\\
&\Rightarrow \ \ \ \ \ x^2_a+Lx_a-2x_P\left(x_P+L\right)=0
\end{aligned}

The above quadratic has two roots that are given by

\left\{ \begin{array}{c}
\displaystyle{x_a=\frac{-L-\sqrt{L^2+8x_P\left(x_P+L\right)}}{2} < 0} \\
\\
\displaystyle{x_a=\frac{-L+\sqrt{L^2+8x_P\left(x_P+L\right)}}{2} > 0} \end{array}
\right.

Keeping the positive root because we know the alpha particle would have to be to the right of the rod, we conclude that the location $x_a$ at which the alpha particle causes the net force on the rod to be zero is equal to

\boxed{x_a=\frac{-L+\sqrt{L^2+8x_P\left(x_P+L\right)}}{2}}

When point $P$ is extremely far away from the rod, the electric field ${\overrightarrow{E}}_P$ from the rod will resemble that of a point charge $Q$. Indeed, if $x_P\gg L$, then we have $x_P-L\approx x_P$ and therefore

\boxed{{\overrightarrow{E}}_P=\frac{kQ}{x_P\left(x_P-L\right)}\ \ \hat{x}\approx \frac{kQ}{x^2_P}\ \ \hat{x}}