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Chapter 16: Problem 7

In Millikan's oil drop experiment, an oil drop of radius \(\mathrm{r}\) and charge \(q\) is held in equilibrium between plates of a parallel plate capacitor when the potential difference is \(\mathrm{V}\). To keep a drop of radius \(2 r\) with charge \(2 q\) in equilibrium between the plates, the potential difference required is (a) \(\mathrm{V}\) (b) \(2 \mathrm{~V}\) (c) \(4 \mathrm{~V}\) (d) \(8 \mathrm{~V}\)

Short Answer

Expert verified
The required potential difference is (d) \(8V\).

Step by step solution

01

Understanding the Forces

In Millikan's oil drop experiment, the oil drop is in equilibrium, meaning the forces acting on it are balanced. The two main forces are the gravitational force, pulling it downwards, and the electric force exerted by the electric field of the capacitor, pushing it upwards when the charges repel.
02

Determine Gravitational Force

The gravitational force can be calculated using the formula for the weight of the oil drop: \( F_g = m g \). The mass \( m \) of the drop is based on its volume \( V \) and density \( \rho \). For a sphere, \( V = \frac{4}{3}\pi r^3 \), providing \( m = \rho V = \rho \cdot \frac{4}{3} \pi r^3 \), so \( F_g = \rho \cdot \frac{4}{3} \pi r^3 g \).
03

Calculate Electric Force

The electric force is given by \( F_e = q E \), where \( E \) is the electric field between the plates of the capacitor. With a potential difference \( V \), the electric field is \( E = \frac{V}{d} \), where \( d \) is the separation between the plates. Thus, \( F_e = q \frac{V}{d} \).
04

Set Up Equilibrium Condition

For equilibrium, the forces must balance such that \( F_g = F_e \). Substitute the expressions: \( \rho \cdot \frac{4}{3} \pi r^3 g = q \frac{V}{d} \).
05

Adjust for New Conditions

For the new drop with radius \( 2r \) and charge \( 2q \), recalculate each force. The new gravitational force becomes \( F_g = \rho \cdot \frac{4}{3} \pi (2r)^3 g = 8 \rho \cdot \frac{4}{3} \pi r^3 g \). The electric force required is \( 2q \frac{V'}{d} \). Set \( 8 \rho \cdot \frac{4}{3} \pi r^3 g = 2q \frac{V'}{d} \).
06

Solve for New Potential Difference

Substitute the equilibrium condition: \( 8 \rho \cdot \frac{4}{3} \pi r^3 g = 2q \frac{V'}{d} \). Recognize \( \rho \cdot \frac{4}{3} \pi r^3 g \) from the original balance equation. Consequently, \( V' = 8V \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Gravitational Force in Millikan's Experiment
In Millikan's oil drop experiment, the gravitational force is pivotal for understanding how oil drops achieve equilibrium. This force is what pulls the oil drop downward towards the Earth. To calculate it, we use the formula for weight:
  • Gravitational Force: \( F_g = mg \)
Here, \( m \) signifies the mass of the oil drop, while \( g \) represents the acceleration due to gravity, approximately \( 9.8 \text{ m/s}^2 \) on Earth's surface. The mass \( m \) can be determined given the volume and density of the oil drop:
  • Volume of the oil drop, for a sphere: \( V = \frac{4}{3}\pi r^3 \)
  • Mass: \( m = \rho V = \rho \cdot \frac{4}{3} \pi r^3 \)
Thus, the gravitational force becomes:
  • \( F_g = \rho \cdot \frac{4}{3} \pi r^3 g \)
This calculation is essential to determine the balance of forces on a charged oil drop.
Electric Force in the System
The electric force in Millikan's experiment acts upward on the oil drop, counteracting the gravitational force. This force arises due to the interaction between the electric field within the capacitor and the charge on the oil drop.
  • Electric Force: \( F_e = qE \)
The electric field \( E \) within a parallel plate capacitor is defined by the potential difference \( V \) across the plates and their separation \( d \):
  • Electric Field: \( E = \frac{V}{d} \)
Consequently, the electric force becomes:
  • \( F_e = q \frac{V}{d} \)
The strength of this force is crucial for holding the oil drop in equilibrium, as it needs to exactly balance the gravitational pull.
Understanding Equilibrium Condition
To maintain equilibrium in Millikan's experiment, the forces acting on the oil drop must be balanced. This principle implies:
  • Equilibrium Condition: \( F_g = F_e \)
By substituting the formulas for gravitational and electric forces, this condition becomes:
  • \( \rho \cdot \frac{4}{3} \pi r^3 g = q \frac{V}{d} \)
This equation highlights that for equilibrium:
  • The strength of the electric force matches the gravitational pull
  • The potential difference \( V \) across the plates directly influences the electric force
Adjustments in charge and radius, such as doubling them, require reconsiderations of \( V \) to maintain balance.
Role of Potential Difference
In Millikan's experiment, the potential difference \( V \) between the capacitor plates is a critical factor. It determines the electric field within the capacitor, which in turn affects the electric force acting on the oil drop.
  • Potential Difference: Determines the magnitude of \( E = \frac{V}{d} \)
An increase in potential difference strengthens the electric field, resulting in a greater electric force:
  • Required adjustments: Changes in \( r \) and \( q \) of the oil drop require recalculating \( V \)
For instance, doubling the oil drop's radius and charge requires reevaluating \( V' \) to maintain balance. Specifically:
  • The needed potential difference to balance doubled parameters results in \( V' = 8V \)
Hence, careful calibration of \( V \) is essential for stable equilibrium in the experiment.

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Most popular questions from this chapter

Two copper spheres of same radii, one hollow and the other solid, are charged to the same potential. Which will hold more charge? (a) Solid sphere (b) Hollow sphere (c) Both will hold equal charge (d) Nothing can be predicted

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The mathematical form of Gauss' law is $$ \varepsilon_{o} \oint \vec{E} \cdot d \vec{S}=q $$ In this reference which of the following is correct? (a) \(E\) depends on the charge \(q\) which is enclosed within the Gaussian surface only (b) \(E\) depends on the charge which is inside and outside the Gaussian surface (c) \(E\) does not depend on the magnitude of charge \(q\) (d) All of the above

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