Animated Solution for Physics - Electric Charges and Fields: An oil drop of radius $2\text{ mm}$ with a density $3{\text{ g cm}}^{-3}$ is held stationary under a constant electric field $3.55\times 10^5{\text{ Vm}}^{-1}$ in the Millikan's oil drop experiment. What is the number of excess electrons that the oil drop will possess? (Take, $g=9.81{\text{ m/s}}^2$)

The Millikan oil drop experiment is one of the most beautiful and elegant experiments in the history of physics. It allowed us to peek into the quantum nature of our universe and measure the fundamental charge of a single electron. In this problem, we are going to step into Robert Millikan's shoes and analyze a stationary oil drop suspended in an electric field.

Imagine a tiny oil drop suspended in mid-air. It is not falling! Why? Because the downward pull of gravity is perfectly balanced by an upward electrical force. For the drop to be stationary, the net force acting on it must be exactly zero.

This means the gravitational force, $F_g=mg$, must equal the electrostatic force, $F_e=qE$. We can write our master equilibrium equation as:

We don't know the mass of the drop directly, but we are given its density $\rho$ and its radius $r$. We know that mass is density times volume. Assuming the oil drop is a perfect sphere due to surface tension, its volume is $V=\frac{4}{3}\pi r^3$. Therefore, the mass is:

Furthermore, the total charge $q$ on the drop is not just a random continuous value. It is quantized! It must be an integer multiple of the elementary charge $e$. So, we can write $q=ne$, where $n$ is the number of excess electrons.

Substituting these expressions back into our equilibrium equation, we get:

We want to find $n$, so let's rearrange the equation:

The Silent Trap

Before we rush into plugging in the numbers, we must be extremely careful with our units. This is where many students make silly mistakes. We need everything in standard SI units.

The radius is given as $2\text{ mm}$, which is $2\times 10^{-3}\text{ m}$. The density is given as $3{\text{ g/cm}}^3$. To convert this to ${\text{kg/m}}^3$, we multiply by $1000$, giving us $3\times 10^3{\text{ kg/m}}^3$.

Now, we carefully plug in our given values into the rearranged equation:

Let's crunch the numbers. The powers of ten will simplify nicely:

After evaluating the expression, we find:

This means there are approximately $1.73\times 10^{10}$ excess electrons on this tiny oil drop. It's a massive number, but remember, electrons are incredibly small! This beautiful balance of forces is exactly how we unlocked the secrets of the quantum world.