An electric field,$\overrightarrow{\mathbf{E}}$with an average magnitude of about$\mathbf{150}\frac{\mathbf{\text{N}}}{\mathbf{\text{C}}}$points downward in the atmosphere near Earth’s surface.We wish to “float” a sulfur sphere weighing$\mathbf{4}\mathbf{.4}\mathbf{\text{N}}$in this field by charging the sphere. (a) What charge (both sign and magnitude) must be used? (b) Why is the experiment impractical?

a) Average magnitude of the electric field $E_{avg}=150\frac{\text{N}}{\text{C}}$.

b) Weight of sulphur sphere $W\left(orF\right)=\text{4.4\hspace{0.17em}N}$.

Using the concept of the electric field, determine both the sign and the magnitude of the charge using the given data. Again, using the same concept, determine the reason for the experiment being impractical.

The electrostatic force acting on the particle is as follows:

$\overrightarrow{F}=q\overrightarrow{E}$ …… (i)

Here, $qq$is the charge of the particle (including its sign) and$\overrightarrow{E}$is the electric field that other charges have produced at the location of the particle.

The magnitude of the electric field is as follows:

$E=\frac{q}{4\pi {\epsilon }_o{R}^2}$ ……. (ii)

Here,

$R$= The distance of field point from the charge

$q$= charge of the particle

To float the sphere, we have to apply upward force that is equal to gravitational force or weight of the sphere. The force on the charge due to electric field is product of electric field and charge. The sign of charge decides the direction of the force. The given electric field is pointing downward direction. Therefore, to create a force in the upward direction, we should have a negative charge on the sphere. The magnitude of the charge is found by working with the absolute value of equation (i) as:

$\begin{array}{c}\left|q\right|=\frac{F}{E}\\ =\frac{4.4\text{N}}{150\text{N}/\text{C}}\\ =0.029\text{C}\end{array}$

Hence, the value of the charge is.$-0.029\text{C}$

Since, the mass of the sphere is given as:

$m=\frac{F}{g}$

Substitute the values and solve as:

$\begin{array}{c}m=\frac{4.4\text{N}}{9.8\text{m}/{\text{s}}^2}\\ =0.45\text{kg}\end{array}$

From the density of sulphur$(2.1\times {10}^3\text{\hspace{0.17em}kg}/{\text{m}}^3)$ ,find the radius of the sphere as follows:

$\begin{array}{c}{\rho }_{sulfur}\left(\frac{4}{3}\pi r^3\right)=m_{sphere}\\ r={\left(\frac{3m_{sphere}}{4\pi {\rho }_{sulfur}}\right)}^{\frac{1}{3}}\\ =0.037\text{m}\end{array}$

Now use the equation (ii), to find the electric field using the above data:

$\begin{array}{c}E=\frac{(9\times {10}^9\text{kg}\cdot {\text{m}}^3/{\text{s}}^2\cdot {\text{C}}^2)\left(0.029\text{C}\right)}{{\left(0.037{\text{m}}^2\right)}^2}\\ =1.90\times {10}^{11}\text{\hspace{0.17em}N}/\text{C}\\ \approx 2\times {10}^{11}\text{\hspace{0.17em}N}/\text{C}\end{array}$

The quantity of the electric field is too large. It will be very difficult to maintain this high electric field in the air. Also, with this large electric field, as the repulsive forces would explode the sphere.

Thus, the large value of the electric field makes the experiment too impractical.