Question

The identical small spheres shown in FIGURE CP22.75 are charged to $+100$ nC and $-100$nC. They hang as shown in a $100,000$ N/C electric field. What is the mass of each sphere?

Short answer

The mass of each sphere is$4.1g$.

Step-by-step solution

Given information

A free-body diagram for this is given below

Explanation

The net force on the given charge$-100\text{nC}$is the sum of the forces coming from the positive charge, gravity, and electric field. By resolving forces in the X direction, we get

$$\begin{gathered} \sum Fx=FE-Fe-T\cos (90-\theta )=0 \\ \sum Fx=FE-Fe-T\cos 80=0\dots \dots \dots \left(1\right) \end{gathered}$$

Resolving forces in the Y direction

role="math" localid="1648482100648" $\sum Fy=T\sin 80-FG=0\dots \dots \left(2\right)$

Here we know that $F_G=mg$

Now equation(2) becomes

$T=\frac{mg}{\sin 80}$

Substitute above value in equation(1)

$$\begin{gathered} FE-Fe=T\cos 80 \\ FE-Fe=\sin 80\times mg/\cos 80 \\ FE-Fe=mg/\tan 80 \end{gathered}$$

We know that

$$\begin{gathered} FE=qE \\ Fe=\frac{1}{4\pi \epsilon 0}\times \frac{qQ}{{\mathbf{r}}^2} \end{gathered}$$

Solving for mass m gives

$$\begin{gathered} qE-\frac{1}{4\pi \epsilon 0}\times \frac{qQ}{{\mathbf{r}}^2}=mg/\tan 80 \\ m=\frac{q\tan 80}{g}\left(E-\frac{Q}{4\pi \epsilon 0{\mathbf{r}}^2}\right) \end{gathered}$$

$$\begin{gathered} m=\frac{100\times {10}^{-9}\times \tan 80}{9.8}\left({10}^5-\left(9\times {10}^{9)}\frac{100\times {10}^{-9}}{{\sin }^{2}10}\right)\right. \\ m=4.1g \end{gathered}$$