What is the electric field strength at a point in space where a proton experiences an acceleration of 2.4 million “g’s”?
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Electric Field
Problem 49a
Textbook Question
Consider an oil droplet of mass m and charge q. We want to determine the charge on the droplet in a Millikan-type experiment. We will do this in several steps. Assume, for simplicity, that the charge is positive and that the electric field between the plates points upward. An electric field is established by applying a potential difference to the plates. It is found that a field of strength E₀ will cause the droplet to be suspended motionless. Write an expression for the droplet's charge in terms of the suspending field E₀ and the droplet's weight mg.

1
Step 1: Begin by identifying the forces acting on the oil droplet. Since the droplet is motionless, the net force acting on it must be zero. The two forces at play are the gravitational force (weight) acting downward, given by \( F_g = mg \), and the electric force acting upward, given by \( F_e = qE_0 \).
Step 2: Apply the condition for equilibrium. Since the droplet is suspended motionless, the upward electric force must exactly balance the downward gravitational force. Mathematically, this can be expressed as \( F_e = F_g \), or \( qE_0 = mg \).
Step 3: Solve for the charge \( q \) on the droplet. Rearrange the equation \( qE_0 = mg \) to isolate \( q \). This gives \( q = \frac{mg}{E_0} \).
Step 4: Interpret the result. The charge \( q \) on the droplet is directly proportional to its weight \( mg \) and inversely proportional to the strength of the electric field \( E_0 \). This means that a stronger electric field would require less charge to balance the droplet's weight.
Step 5: Conclude by noting that this expression \( q = \frac{mg}{E_0} \) provides a way to calculate the charge on the droplet if the mass \( m \), gravitational acceleration \( g \), and electric field strength \( E_0 \) are known.

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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Electric Field
An electric field is a region around a charged object where other charged objects experience a force. It is defined as the force per unit charge and is represented by the symbol E. In the context of the Millikan experiment, the electric field is created between two plates by applying a potential difference, allowing charged droplets to be manipulated and suspended against gravitational forces.
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Weight of an Object
The weight of an object is the force exerted on it due to gravity, calculated as the product of its mass (m) and the acceleration due to gravity (g). In this scenario, the weight acts downward on the oil droplet, counteracting the upward force exerted by the electric field. Understanding the balance between these forces is crucial for determining the charge on the droplet.
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Force Balance in Millikan Experiment
In a Millikan-type experiment, the charge on a droplet can be determined by analyzing the forces acting on it. When the droplet is suspended motionless, the upward electric force (qE₀) equals the downward gravitational force (mg). This balance allows us to derive the expression for the droplet's charge (q) in terms of the electric field strength (E₀) and the droplet's weight (mg), leading to the equation q = mg/E₀.
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