Hanging Mass!

a) Use Newton’s Second Law to relate the variables and solve for the unknown charge.

b) Using the equations obtained in part (b), solve for the angle.

a) At equilibrium, Newton’s Second Law states:

\begin{equation*}
\sum \vec{F}=\vec{0}.
\end{equation*}

For the \({y-}\)axis, we have:

\begin{equation*}
T \cos \theta – mg = 0,
\end{equation*}

and for the \({x-}\)axis, we have:

\begin{equation*}
q E – T \sin \theta = 0.
\end{equation*}

For the \(2 \times 2\) system of equations, solving for \(q\), we get:

\begin{equation*}
q = \frac{mg \tan \theta}{E},
\end{equation*}

which, with numerical values, gives us:

\begin{equation*}
q \approx 3.9 \times 10^{-5} \, \text{C}.
\end{equation*}

b) Solving for the angle, we get:

\begin{equation*}
\tan \theta = \frac{qE}{mg},
\end{equation*}

where, by applying the arctan function to both sides of the last equation, we obtain:

\begin{equation*}
\theta \approx 36^\circ.
\end{equation*}

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

a) We need to calculate the charge of the hanging mass. To approach this problem, we’ll identify all the forces exerted on the mass \(m\) and apply Newton’s second law to solve for the unknowns. We assume that the system is at equilibrium, so the acceleration \(\vec{a}\) is zero and Newton’s second law can be written as

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

\begin{equation}
\label{newton}
\sum \vec{F}=\vec{0},
\end{equation}

where \(\sum \vec{F}\) is the sum of all the forces exerted on the mass \(m\).

To identify the forces and their directions, we draw a free-body diagram like the one shown in figure 1.

Figure 1: Free-body diagram for the charge \(q\). The forces exerted on the mass are the weight \(\vec{W}\), the electric force \(\vec{F}_e\), and the tension \(\vec{T}\), which decomposes along the positive Y-axis and negative X-axis according to the angle \(\theta\). Notice that since the charge is positive, the electric force is in the same direction of the electric field \(\vec{E}\).

Thus, we can write equation \eqref{newton} as

\begin{equation}
\label{newton2}
\vec{T}+\vec{W}+\vec{F}_e=\vec{0},
\end{equation}

where \(\vec{T}\) is the tension exerted by the rope, \(\vec{W}\) is the weight of the mass, and \(\vec{F}_e\) is the electric forces excerted on the mass, which we assume is positively charged.

The weight force is directed towards the negative Y axis and has a magnitude equal to \(mg\), so we can write

\begin{equation}
\label{weight}
\vec{W}=-mg\,\hat{\textbf{j}}.
\end{equation}

The tension \(\vec{T}\) has magnitude \(T\), and its direction is given by the angle \(\theta\). Using trigonometry, we can write

\begin{equation}
\label{T}
\vec{T}=-T\sin(\theta)\,\hat{\textbf{i}}+T\cos(\theta)\,\hat{\textbf{j}}.
\end{equation}

Finally, the electric force is in the same direction of the electric field in this case because the charge \(q\) is positive. Its magnitude is \(qE\), where \(E\) is the magnitude of the electric field

\begin{equation}
\label{eforce}
\vec{F}_e=qE\,\hat{\textbf{i}}.
\end{equation}

Putting the expressions for the forces given in equations \eqref{weight}, \eqref{T}, and \eqref{eforce} into equation \eqref{newton2}, we obtain

\begin{equation}
\label{newton3}
-T\sin(\theta)\,\hat{\textbf{i}}+T\cos(\theta)\,\hat{\textbf{j}}-mg\,\hat{\textbf{j}}+qE\,\hat{\textbf{i}}=\vec{0},
\end{equation}

which can be written as

\begin{equation}
\label{newton4}
\left(-T\sin(\theta)+qE\right)\,\hat{\textbf{i}}+\left(T\cos(\theta)-mg\right)\,\hat{\textbf{j}}=\vec{0}.
\end{equation}
For the equation above to be consistent, the components X and Y must be zero separately. Focusing on the magnitude of the components along the X axis in equation \eqref{newton4}, we have

\begin{equation}
\label{Fx}
-T\sin(\theta)+qE=0,
\end{equation}

while for the Y axis, we obtain

\begin{equation}
\label{Fy}
T\cos(\theta)-mg=0.
\end{equation}

From equation \eqref{Fy}, we can solve for \(T\) to get

\begin{equation}
\label{tension}
T=\frac{mg}{\cos(\theta)}.
\end{equation}

Using this result in equation \eqref{Fx}, we get

\begin{equation}
-\frac{mg}{\cos(\theta)}\sin(\theta)+qE=0,
\end{equation}

which simplifies to

\begin{equation}
\label{fy2}
-mg\tan(\theta)+qE=0.
\end{equation}

Solving for \(q\) in equation \eqref{fy2}, we have

\begin{equation}
q=\frac{mg\tan(\theta)}{E}.
\end{equation}

Using the numerical values in SI units (\(m=6\times10^{-3}\,\text{kg}\) and \(g=9.81\,\text{m/s}^2\)), we obtain

\begin{equation}
q=\frac{(6\times10^{-3}\,\text{kg})(9.81\,\text{m/s}^2)\tan(20^{\circ})}{550\,\text{N/C}},
\end{equation}

\begin{equation}
q\approx3.9\times 10^{-5}\,\text{C}.
\end{equation}

b) If the electric field is now two times larger, that is \(E=1100\,\text{N/C}\), and the charge is the same as before, we can solve for the new angle \(\theta\) from equation \eqref{fy2} to get

\begin{equation}
mg\tan(\theta)=qE,
\end{equation}

\begin{equation}
\tan(\theta)=\frac{qE}{mg},
\end{equation}

\begin{equation}
\theta=\arctan\left(\frac{qE}{mg}\right),
\end{equation}

Using the numerical values given by the problem and the ones calculated in part (a), we obtain

\begin{equation}
\theta=\arctan\left(\frac{(3.9\times 10^{-5}\,\text{C})(1100\,\text{N/C})}{(6\times10^{-3}\,\text{kg})(9.81\,\text{m/s}^2)}\right),
\end{equation}

\begin{equation}
\theta\approx
36^{\circ}.
\end{equation}

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