A tiny spherical oil drop carrying a net charge q is balanced in still air with a vertical uniform e

Q: Why do we use Stokes' Law here?

Because the oil drop is a tiny sphere falling through a viscous fluid (air). Stokes' Law perfectly describes the viscous drag force on small spherical objects moving at low speeds.

Q: What happens to the buoyant force?

In classic Millikan oil drop problems, the buoyant force of air is often neglected because the density of air is extremely small compared to the density of the oil. However, if we were to be perfectly rigorous, we would use an effective weight $m_{eff}g = \frac{4}{3}\pi r^3 (\rho_{oil} - \rho_{air})g$. Given the data, neglecting it is the intended path.

Q: Why does the drop reach a terminal velocity?

As the drop falls, it accelerates due to gravity. However, the viscous drag force increases with velocity. Eventually, the drag force equals the gravitational force, resulting in zero net force and a constant terminal velocity.

Q: Can we find the mass directly without finding the radius?

No, because the mass depends on the volume, which requires the radius. The terminal velocity gives us a way to find the radius, which then unlocks the mass.

Q: Why is the electric field given with a $\pi$ in it?

It's a classic problem-setter's trick! The $\pi$ in the electric field is designed to perfectly cancel out the $\pi$ from Stokes' Law and the volume of the sphere, making the final calculation surprisingly clean.

Question

A tiny spherical oil drop carrying a net charge $q$ is balanced in still air with a vertical uniform electric field of strength $\frac{81\pi}{7} 	imes 10^5 	ext{ Vm}^{-1}$. When the field is switched off, the drop is observed to fall with terminal velocity $2 	imes 10^{-3} 	ext{ ms}^{-1}$. Given $g = 9.8 	ext{ ms}^{-2}$, viscosity of the air $= 1.8 	imes 10^{-5} 	ext{ Ns m}^{-2}$ and the density of oil $= 900 	ext{ kg m}^{-3}$, the magnitude of $q$ is

Options:

(A) $1.6 	imes 10^{-19} 	ext{ C}$

(B) $3.2 	imes 10^{-19} 	ext{ C}$

(C) $4.8 	imes 10^{-19} 	ext{ C}$

(D) $8.0 	imes 10^{-19} 	ext{ C}$

Sigma Percentile · Accepted Answer

### The Millikan Oil Drop Experiment

Imagine a tiny oil drop suspended in mid-air. It's perfectly balanced because the upward electric force exactly cancels gravity. But when we turn the field off, it starts to fall, eventually reaching a constant terminal velocity due to air drag. Let's analyze both these states to uncover the charge on the drop.

### Balancing Act: The Electric Force

In the first case, the drop is stationary. This means the net force is zero. The upward electric force, $qE$, must be exactly equal to the downward gravitational force, $mg$.

$$qE = mg$$

### The Free Fall: Stokes' Law

Now, when the field is switched off, the drop falls. As it speeds up, the viscous drag force increases until it balances gravity. At this terminal velocity, Stokes' Law tells us that the drag force, $6\pi\eta r v$, equals $mg$.

$$6\pi\eta r v = mg$$

We don't know the mass or the radius of the drop directly. But we know it's a sphere. So, its mass is its volume, $\frac{4}{3}\pi r^3$, multiplied by the density of the oil, $ho$.

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

Let's substitute this mass into our terminal velocity equation. Notice how one $r$ cancels out on both sides. We can rearrange this to solve for the radius squared, and then take the square root to find $r$.

$$6\pi\eta r v = \left(\frac{4}{3}\pi r^3 hoight) g$$

$$r = \sqrt{\frac{9\eta v}{2ho g}}$$

### Unveiling the Charge

Now that we have an expression for the radius, let's plug it back into the Stokes' Law equation to get a complete expression for the weight, $mg$, entirely in terms of known quantities.

$$mg = 6\pi\eta v \sqrt{\frac{9\eta v}{2ho g}}$$

Remember our very first equation? The charge $q$ is simply $mg$ divided by the electric field $E$. Let's substitute our massive expression for $mg$ into this. This is our master equation for the charge.

$$q = \frac{6\pi\eta v}{E} \sqrt{\frac{9\eta v}{2ho g}}$$

Don't get intimidated by the algebra. It's time to plug in all the given values. We have the electric field, terminal velocity, viscosity, density, and gravity. Carefully substitute them into the master equation.

$$q = \frac{6\pi(1.8 	imes 10^{-5})(2 	imes 10^{-3})}{\frac{81\pi}{7} 	imes 10^5} \sqrt{\frac{9(1.8 	imes 10^{-5})(2 	imes 10^{-3})}{2(900)(9.8)}}$$

Let's simplify the terms. The term inside the square root simplifies beautifully to $\frac{3}{7} 	imes 10^{-5}$. Multiplying everything out, the $\pi$s cancel, the $7$s cancel, and we are left with a neat $8.0 	imes 10^{-19} 	ext{ C}$.

$$q = 8.0 	imes 10^{-19} 	ext{ C}$$

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Visualized Solution

$Physical Setup$

$Equilibrium Condition$

$Terminal Velocity Condition$

$Mass of the Drop$

$Finding the Radius$

$Expressing m g$

$Charge Equation$

$Substitution$

$Final Calculation$

The Sigma Insight: Electric Field

The Millikan Oil Drop Experiment

Balancing Act

The Free Fall

Unveiling the Charge

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Physical Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Equilibrium Condition

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Terminal Velocity Condition

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Mass of the Drop

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Finding the Radius

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Expressing mg

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Charge Equation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substitution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Calculation