Two large circular discs separated by a distance of 0.01 m are connected to a battery via a switch a

Q: Why does the electric force act upwards?

The oil drop has excess electrons, making it negatively charged. The top plate is connected to the positive terminal (implied by the setup to make it float), so the electric field points downwards, and the force on a negative charge is opposite to the field, i.e., upwards.

Q: What if we didn't neglect buoyancy?

If buoyancy was considered, the upward forces would be $qE + F_b = mg$, where $F_b$ is the buoyant force from the displaced air. This would mean a slightly smaller electric force is needed to balance the drop.

Q: Why do we use the formula for the volume of a sphere?

Surface tension pulls small liquid drops into a spherical shape to minimize surface area, so we can safely assume the oil drop is a perfect sphere.

Q: What does terminal velocity have to do with this?

The problem mentions terminal velocity to indicate that before the switch was closed, the drop was falling at a constant speed due to air drag balancing gravity. Once the switch is closed, it stops completely, meaning the new electric force replaces drag to balance gravity.

Q: Can the number of electrons be a fraction?

No! Charge is quantized. You can only have an integer number of electrons. If your calculation gives 5.98, it means the answer is exactly 6.

Question

Two large circular discs separated by a distance of $0.01 	ext{ m}$ are connected to a battery via a switch as shown in the figure. Charged oil drops of density $900 	ext{ kg m}^{-3}$ are released through a tiny hole at the center of the top disc. Once some oil drops achieve terminal velocity, the switch is closed to apply a voltage of $200 	ext{ V}$ across the discs. As a result, an oil drop of radius $8 	imes 10^{-7} 	ext{ m}$ stops moving vertically and floats between the discs. The number of electrons present in this oil drop is _____. (neglect the buoyancy force, take acceleration due to gravity $= 10 	ext{ ms}^{-2}$ and charge on an electron $(e) = 1.6 	imes 10^{-19} 	ext{ C}$)

Sigma Percentile · Accepted Answer

The problem presents a beautiful recreation of the famous Millikan oil drop experiment, a cornerstone of modern physics that first measured the elementary charge of an electron. Let's dive into the mechanics of this floating drop!

### Analyzing the Setup

Imagine a tiny oil drop falling through a hole in the top disc. Initially, it falls under the influence of gravity, eventually reaching a constant terminal velocity due to air resistance.

However, the moment we close the switch, a $200 	ext{ V}$ potential difference is applied across the discs. This creates a strong, uniform electric field between them. Suddenly, the drop stops moving and hovers perfectly in mid-air. This tells us that a new upward force has perfectly canceled out the downward pull of gravity.

### The Master Equation

First, we need to determine the strength of this invisible electric field. The electric field $E$ between two parallel plates is given by the voltage $V$ divided by the distance $d$:

$$E = \frac{V}{d}$$

Substituting our given values:

$$E = \frac{200}{0.01} = 2 	imes 10^4 	ext{ V/m}$$

For the drop to float, the upward electric force must balance the downward gravitational force. This gives us our master equilibrium equation:

$$qE = mg$$

### Unpacking the Variables

We don't know the charge $q$ or the mass $m$ directly. However, we know that charge is quantized. The total charge $q$ is simply the number of excess electrons $n$ multiplied by the elementary charge $e$:

$$q = ne$$

The mass $m$ of the spherical oil drop can be found using its density $ho$ and volume $V$:

$$m = ho 	imes \frac{4}{3}\pi r^3$$

Substituting these into our master equation, we get:

$$neE = ho \frac{4}{3}\pi r^3 g$$

### Final Calculation

Now, we substitute all the known values into this expanded equation:

$$n 	imes 1.6 	imes 10^{-19} 	imes 2 	imes 10^4 = 900 	imes \frac{4}{3}\pi (8 	imes 10^{-7})^3 	imes 10$$

This looks intimidating, but it's just a matter of careful arithmetic. Let's isolate $n$:

$$n 	imes 3.2 	imes 10^{-15} \approx 900 	imes 4.19 	imes 512 	imes 10^{-21} 	imes 10$$

$$n 	imes 3.2 	imes 10^{-15} \approx 1.93 	imes 10^{-14}$$

$$n \approx \frac{1.93 	imes 10^{-14}}{3.2 	imes 10^{-15}} \approx 6$$

**The drop has exactly 6 excess electrons!** This elegant balance of forces allows us to count individual electrons, showcasing the profound beauty of physics.

Enter Numerical Value:

Visualized Solution

\text{The Setup}

\text{Electric Field Formula}

\text{Substituting Values}

\text{Calculating Electric Field}

\text{Equilibrium Condition}

\text{Expanding Charge and Mass}

\text{Substituting All Values}

\text{Solving for } n

\text{Final Answer}

\text{Conclusion}

The Sigma Insight: Electric Field

Analyzing the Setup

The Master Equation

Unpacking the Variables

Final Calculation

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