Answer to Question #138563 in Electric Circuits for Sanamtha

Question

Two plane parallel conducting plates 1.50 cm apart are held horizontal, one above the other in air. The upper plate is maintained at a positive potential of 1500 volts while the other lower plate is earthed. (a) Calculate the number of electrons which must be attached to a small oil drop of mass 4.90 x 10-12g, if it remains stationary in the air between the plates. (Assume that the density of the air is negligible in comparison with that of oil.) (b) If the potential of the upper plate is suddenly changed to -1500 volts, what is the initial acceleration of the charge drop? Indicate, giving reasons, how the acceleration will change.

Answer 1

Explanations & Calculations

• It is known that the electric field strength between two plates is in a uniform nature & that can be represented by $\small E=\large \frac{\Delta V}{d}$ where d is the separation between the plates.
• Therefore the electric field strength in this region is

$\qquad\qquad \begin{aligned} \small E &= \small \frac{(1500-0)V}{0.015m}\\ &=\small10^5NC^{-1} \end{aligned}$

• Then the drop of oil is to be balanced under this electric field. And if the needed number of electrons are n then

$\qquad\qquad \begin{aligned} \small F=E(nq)&=mg\\ \small n&= \small \frac{4.9\times10^{-15}kg\times9.8ms^{-2}}{10^5NC^{-1}\times(1.6\times10^{-19}C)}\\ &= \small \bold{3} \end{aligned}$

• After the voltage is reversed to -1500V the electric field acts in the reversed direction with the same magnitude. And the magnitude is just equal to mg as calculated above. Therefore the net force downward the drop of oil is doubled. Hence,

$\qquad\qquad \begin{aligned} \small F&=\small ma\\ \small E(nq)+mg&= \small ma\\ \small mg+mg&= \small ma\\ \small a&= \small 2g=\bold{19.6ms^{-2}} \end{aligned}$

• Since E does not change over the region between the plates, the net force remains constant. Therefore no any change of the acceleration observed during the motion within this region.