Answer
b. Make the rod $ $ $SHORTER$ $ $ without changing the charge.
e. $ADD$ $ $ charge to the rod.
h. Move the dot $ $ $CLOSER$ $ $ to the rod.
Options B, E, H
Work Step by Step
The equation for the electric field at a point perpendicular to the rod's long side is $\quad\vec{E}_{rod} = \displaystyle \frac{kQ}{r\sqrt{r^2+\frac{L^2}{4}}}\qquad$
where...
$r$ is the distance between the rod and the point,
$L$ is the length of the rod, $Q$ is the charge of the rod, and
$k$ is the electrostatic constant: $\displaystyle \frac{1}{4\pi\epsilon_0}$.
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$\textbf{CORRECT: b. Make the rod $ $ $SHORTER$ $ $ without changing the charge.}$
CORRECT: b. Make the rod $ $ $SHORTER$ $ $ without changing the charge.
If we make the rod shorter, then $L$ decreases. Since $L$ is in the denominator of the equation, $\vec{E}_{rod}$ is inversely proportional to $L$ such that $\displaystyle \vec{E}_{rod} \propto \frac{1}{L}$. Therefore, decreasing the magnitude of $L$ will strengthen the electric field.
$\textbf{INCORRECT: c. Make the rod wider without changing the charge}$
The electric field strength produced by a thin charged rod is independent of its width.
Since $r$ is measured from the center of the rod, the thickness of the rod doesn't effect the electric field strength.
Please note that a rod's linear charge density is $\displaystyle \lambda=\frac{Q}{L}$, there is no term for the thickness of the rod.
$\textbf{CORRECT: e. $ADD$ $ $ charge to the rod.}$
Since the charge of the rod, $Q$ is in the numerator of the equation, the electric field and charge are directly related: $\vec{E}_{rod} \propto Q$. Therefore, increasing (adding) charge to the rod will strengthen the electric field.
$\textbf{CORRECT: h. Move the dot $ $ $CLOSER$ $ $ to the rod.}$
As said previously, decreasing the distance between the point and the rod will strengthen the electric field.
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Therefore:
b. Make the rod $ $ $SHORTER$ $ $ without changing the charge.
e. $ADD$ $ $ charge to the rod.
h. Move the dot $ $ $CLOSER$ $ $ to the rod.
Options B, E, H
