Hovering Mass!

Use Newton’s Second Law and the electrical force to get the mass.

Newton’s Second Law states:

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

where, at equilibrium, \(a=0\). Then, by using the fact that the electrical force \(F = qE\), and by applying Newton’s Second Law along the \({y-}\)axis, we get:

\begin{equation*}
qE – mg=0.
\end{equation*}

Solving for \(m\):

\begin{equation*}
m = \frac{qE}{g},
\end{equation*}

which, with numerical values, gives us:

\begin{equation*}
m \approx 3.67 \times 10^{-3} \, \text{kg}.
\end{equation*}

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

The problem asks us to find the mass of the object. Let’s start by writing

\begin{equation}
\label{efield}
\vec{E}=E\,\hat{\textbf{k}},
\end{equation}

where \(E\) is the magnitude of the electric field, a quantity given by the prompt.

If we want the object to be at equilibrium, we apply Newton’s second law and demand that the acceleration \(\vec{a}\) is zero. Newton’s second law reads,

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

where \(\sum \vec{F}\) is the sum of all forces exerted on the object. The relevant forces are the weight \(\vec{W}\) and the electric force \(\vec{F}_e\), as seen in figure 1.

Figure 1: Free-body diagram for the object with the electric force \(\vec{F}_{E}\) upwards in the same direction as the electric field \(\vec{E}\) and the weight \(\vec{W}\) downwards, along the negative Y-axis.

Then, equation \eqref{newton} is:

\begin{equation}
\label{newton2}
\vec{W}+\vec{F}_e=m\vec{a}.
\end{equation}

Since the object is at equilibrium, we can demand the acceleration to be zero in equation \eqref{newton2}; hence, we get

\begin{equation}
\label{newton3}
\vec{W}+\vec{F}_e=\vec{0}.
\end{equation}

The electric force can be expressed in terms of the electric field \(\vec{E}\) and the charge of the object \(q\) as

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

Using the result of equation \eqref{efield} into equation \eqref{eforce}, we obtain

\begin{equation}
\label{eforce2}
\vec{F}_e=qE\,\hat{\textbf{k}}.
\end{equation}

Furthermore, as seen in the figure, the force of weight \(\vec{W}\) points towards the negative Z axis, so we can write

\begin{equation}
\label{weight}
\vec{W}=-W\,\hat{\textbf{k}},
\end{equation}

where \(W\) is the magnitude of the weight, given by the multiplication of the mass of the object \(m\) by the gravitational acceleration of earth \(g\). Then, equation \eqref{weight} becomes

\begin{equation}
\label{weight2}
\vec{W}=-mg\,\hat{\textbf{k}}.
\end{equation}

Using the explicit expressions for the electric force and weight given in equations \eqref{eforce2} and \eqref{weight2} into equation \eqref{newton3}, we obtain

\begin{equation}
\label{newton4}
-mg\,\hat{\textbf{k}}+qE\,\hat{\textbf{k}}=\vec{0}.
\end{equation}

Focusing on the vector components in equation \eqref{newton4} we arrive to the following expression

\begin{equation}
-mg+qE=0.
\end{equation}

Solving for \(m\), we get

\begin{equation}
m=\frac{qE}{g}.
\end{equation}

Using the numerical values for each variable, we can write

\begin{equation}
m=\frac{(30\,\mu\text{C})(1200\,\text{N/C})}{9.81\,\text{m/s}^2},
\end{equation}

\begin{equation}
m\approx 3.67\times10^{-3}\,\text{kg}=3.67\,\text{g}.
\end{equation}

You need to be registered and logged in to take this quiz. Log in or Register