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In a Millikan oil-drop experiment (Section \(22-8\) ), a uniform electric field of \(1.92 \times 10^{5} \mathrm{~N} / \mathrm{C}\) is maintained in the region between two plates separated by \(1.50 \mathrm{~cm}\). Find the potential difference between the plates.

Short Answer

Expert verified
The potential difference is 2880 V.

Step by step solution

01

Understand the Problem

We need to find the potential difference between two plates with a given electric field strength and separation distance. This is a direct application of the relationship between electric field, potential difference, and distance in uniform electric fields.
02

Identify the Relevant Formula

The relationship between the electric field (E), potential difference (V), and distance (d) is given by the formula:\[ V = E \cdot d \]where V is the potential difference, E is the electric field strength, and d is the separation between the plates.
03

Calculate the Potential Difference

Substitute the given values into the formula. Here, E = 1.92 \times 10^{5} \, \text{N/C} and d = 1.50 \times 10^{-2} \, \text{m} (converted from cm to m). Plugging these into the formula gives:\[ V = (1.92 \times 10^{5} \, \text{N/C}) \times (1.50 \times 10^{-2} \, \text{m}) \]Calculate to find that \( V = 2880 \, \text{V} \).

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Key Concepts

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

Millikan oil-drop experiment
The Millikan oil-drop experiment is a crucial historical experiment in physics developed by Robert Millikan and Harvey Fletcher. It was designed to measure the elementary electric charge of an electron. The beauty of this experiment lies in its simplicity and effectiveness. It involved observing tiny oil droplets between two metal plates and adjusting the voltage to balance the gravitational and electric forces acting on the droplets. This balance allowed Millikan to determine the charge of an individual electron, establishing that electric charge is quantized. Today, we can appreciate this experiment as a pioneering study that helped confirm the discrete nature of electric charge.
Uniform electric field
In physics, a uniform electric field is one where the electric field strength is constant at every point in the region. Between two parallel plates, this type of field exists when the plates are close and the area they cover is large relative to the separation distance. The field lines run parallel to one another, indicating that the force exerted on a charge is consistent throughout the field. Uniform electric fields make calculations of force, potential difference, and other properties simple with straightforward formulas. This consistency is why they are often used in experiments, like the Millikan oil-drop experiment, which helps us understand fundamental electrical properties.
Electric field strength
Electric field strength, denoted as E, is a measure of the force per unit charge that a charged particle would experience in an electric field. It is a vector quantity, meaning it has both magnitude and direction, typically represented in units of newtons per coulomb (N/C). In the context of two charged plates, the electric field strength is uniform and provides an indirect measure of how much potential energy a charge would have when placed between the plates. Understanding this strength is essential to calculating potential differences and predicting the behavior of charged particles in various fields. The constant field strength in the exercise helps establish the potential difference quickly with the formula \( V = E \cdot d \).
Plate separation distance
The plate separation distance refers to the physical distance between the two parallel plates in a capacitor setup, which directly influences the electric field and potential difference. In the context of uniform electric fields, this distance is a critical factor: a smaller distance results in a stronger electric field for the same voltage, and vice versa. This distance, denoted as \( d \), usually has units of meters, and it plays a significant role in calculating the potential difference, which is the product of the electric field strength and the distance (\( V = E \cdot d \)). The exercise demonstrates how important precise measurements and conversions can be, as even small differences in distance can significantly impact the electric potential. This principle is widely applied in designing capacitors and other electrical components. Regular understanding of how separation impacts field strength allows for better control and application in various electromechanical systems.

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Most popular questions from this chapter

The electric field in a region of space has the components \(E_{y}=\) \(E_{z}=0\) and \(E_{x}=(4.00 \mathrm{~N} / \mathrm{C}) x .\) Point \(A\) is on the \(y\) axis at \(y=3.00 \mathrm{~m}\) and point \(B\) is on the \(x\) axis at \(x=4.00 \mathrm{~m}\). What is the potential difference \(V_{B}-V_{A} ?\)

Figure 24-32 shows a rectangular array of charged particles fixed in place, with distance \(a=39.0 \mathrm{~cm}\) and the charges shown as integer multiples of \(q_{1}=3.40 \mathrm{pC}\) and \(q_{2}=6.00 \mathrm{pC}\). With \(V=0\) at infinity, what is the net electric potential at the rectangle's center? (Hint: Thoughtful examination can reduce the calculation.)

What is the magnitude of the electric field at the point \((3.00 \hat{\mathrm{i}}-2.00 \hat{\mathrm{j}}+4.00 \hat{\mathrm{k}}) \mathrm{m}\) if the electric potential is given by \(V=\) \(2.00 x y z^{2}\), where \(V\) is in volts and \(x, y\), and \(z\) are in meters?

Two metal spheres, each of radius \(3.0 \mathrm{~cm}\), have a center-to-center separation of \(2.0 \mathrm{~m} .\) Sphere 1 has charge \(+1.0 \times\) \(10^{-8} \mathrm{C} ;\) sphere 2 has charge \(-3.0 \times 10^{-8} \mathrm{C}\). Assume that the separation is large enough for us to say that the charge on each sphere is uniformly distributed (the spheres do not affect each other). With \(V=0\) at infinity, calculate (a) the potential at the point halfway between the centers and the potential on the surface of (b) sphere 1 and (c) sphere 2 .

If Earth had a uniform surface charge density of \(1.0\) electron/m \(^{2}\) (a very artificial assumption), what would its potential be? (Set \(V=0\) at infinity.) What would be the (b) magnitude and (c) direction (radially inward or outward) of the electric field due to Earth just outside its surface?

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