Two very large horizontal sheets are 4.25 cm apart and carry equal but opposite uniform surface charge densities of magnitude \(\sigma \). You want to use these sheets to hold stationary in the region between them an oil droplet of mass 486 mg that carries an excess of five electrons. Assuming that the drop is in vacuum, (a) which way should the electric field between the plates point, and (b) what should \(\sigma \) be?

Given,

The distance between the two horizontal sheets \(d = 4.15cm\), an oil droplet of mass \(m = 486\mu g\), the oil droplet has a charge \(q = 5e\), as shown in the figure below.

Since, the two plates are used to hold the droplet, the force of the electric field is upwards against the gravitational force. As the charge of the oil droplet is negative implying that the electric field if downward to apply an electric force in the opposite direction.

Now, to hold the oil droplet between the two plates, the electric force must be equal to the weight of the oil droplet.

That is: \(qE = mg\) …………………\(\left( i \right)\)

Where, \(q\) is the charge of the oil droplet, m is the mass of the oil droplet and \(E\) is the electric field produced between the two large sheets.

Again, the electric field between the two sheets is given by: \(E = \frac{\sigma }{{{\varepsilon _ \circ }}}\).

Where, \({\varepsilon _ \circ }\)is the electric constants whose value is . Now, substituting the value of \(E\) in equation \(\left( i \right)\) and solving for s we get:

\(\begin{aligned}qE = mg\\(5e)\frac{\sigma }{{{\varepsilon _ \circ }}} = mg\\\sigma = {\varepsilon _ \circ }\frac{{mg}}{{5e}}\\ = (8.85x{10^{ - 12}}{C^2}/N.{m^2})\frac{{(486 \times {{10}^{ - 9}}kg)(9.8m/{s^2})}}{{5(1.60 \times {{10}^{ - 19}}}}\\ = 52.7C/{m^2}\end{aligned}\)

Therefore, the surface charge distance \(\sigma \)is \(52.7C/{m^2}\)