Question

A negatively charged droplet has a mass of \(5 \times 10^{-16} \mathrm{~kg}\) and carries a charge of \(8 \times 10^{-18} \mathrm{C}\). The droplet falls through the hole in the upper plate when the electric field is \(0 \mathrm{~V} / \mathrm{m}\). How does the drop move within the plates as the electric field is increased slowly from \(0 \mathrm{~V} / \mathrm{m}\) to \(800 \mathrm{~V} / \mathrm{m}\) ? (Note: Assume that the drop remains between the plates at all times.) 1\. It moves downwards, stops, then moves upwards. 2\. It moves downwards, accelerating all the time. 3\. It moves downwards, stops, then moves downwards again. 4\. It moves downwards, stops, and remains stationary.

Short answer

1. It moves downwards, stops, and then moves upwards.

Step-by-step solution

Identify the Forces Acting on the Droplet

There are two main forces acting on the droplet: 1. Gravitational force (downward direction): F_gravity = m * g. Where m is the mass of the droplet and g is the acceleration due to gravity (approximately 9.81 m/s²). 2. Electric force (upward direction, since the droplet is negatively charged): F_electric = q * E. Where q is the charge of the droplet and E is the electric field strength.

Determine the Droplet's Motion at Zero Electric Field

When the electric field is zero, there is no electric force acting on the droplet. Therefore, the droplet will only experience a downward gravitational force, which will cause it to fall through the hole in the upper plate.

Calculate the Gravitational Force Acting on the Droplet

To calculate the gravitational force, we use the formula: F_gravity = m * g Given, m = \(5 \times 10^{-16} kg\) g = \(9.81 m/s^2\) F_gravity = (5 x \(10^{-16} kg\)) * (9.81 \(m/s^2\)) F_gravity = \(4.905 \times 10^{-15} N\)

Compute the Maximum Electric Force

To compute the maximum electric force when the electric field is 800 V/m, we use the formula: F_electric = q * E Given, q = \(8 \times 10^{-18} C\) E = \(800 V/m\) F_electric = (8 x \(10^{-18} C\)) * (800 \(V/m\)) F_electric = \(6.4 \times 10^{-15} N\)

Compare Gravitational and Electric Forces and Determine the Droplet's Motion

Now that we have the gravitational force (F_gravity = \(4.905 \times 10^{-15} N\)) and the maximum electric force (F_electric = \(6.4 \times 10^{-15} N\)), we can determine the droplet's motion. Initially, the droplet will be influenced by gravity and move downwards. As the electric field increases, the upward electric force becomes stronger. When Electric force is equal to gravitational force (F_electric = F_gravity), the droplet will stop moving (no net force on it). Finally, as the electric force keeps increasing and exceeds the gravitational force (F_electric > F_gravity), the droplet will experience a net upward force and move upwards.

Select the Correct Answer

From the analysis above, we can conclude that the droplet moves downwards, stops, and then moves upwards, which corresponds to option 1: 1. It moves downwards, stops, and then moves upwards.