In a photocell, ultraviolet (UV) light provides enough energy to some electrons in barium metal to eject them from the surface at high speed. To measure the maximum energy of the electrons, another plate above the barium surface is kept at a negative enough potential that the emitted electrons are slowed down and stopped, and return to the barium surface. See Fig. 23–52. If the plate voltage is -3.02 V (compared to the barium) when the fastest electrons are stopped, what was the speed of these electrons when they were emitted?
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Electric Potential
Problem 41
Textbook Question
(III) Suppose the flat circular disk of Fig. 23–15 (Example 23–10) has a nonuniform surface charge density that depends on the distance R from the center of the disk: σ = aR². Find the potential V (x) at points along the x axis, relative to V = 0 at x = ∞.

1
Understand the problem: The goal is to find the electric potential V(x) at points along the x-axis due to a flat circular disk with a nonuniform surface charge density σ = aR², where R is the radial distance from the center of the disk. The potential is relative to V = 0 at x = ∞.
Set up the expression for the electric potential: The potential at a point due to a surface charge distribution is given by V = (1 / (4πε₀)) ∫(σ / r) dA, where r is the distance from the charge element to the point of interest, and dA is the infinitesimal area element on the disk.
Express the surface charge density and area element: Substitute σ = aR² into the integral. The infinitesimal area element in polar coordinates is dA = 2πR dR. The distance r from a point on the disk to a point on the x-axis is r = √(x² + R²).
Write the integral for the potential: Substitute σ and dA into the potential formula. The integral becomes V(x) = (1 / (4πε₀)) ∫[aR² / √(x² + R²)] (2πR dR), where the limits of integration are from R = 0 to R = R₀ (the radius of the disk).
Simplify and solve the integral: Factor out constants (a, 2π, and 1 / (4πε₀)) from the integral. The remaining integral is ∫[R³ / √(x² + R²)] dR. Use substitution or a standard integral table to evaluate this expression. The result will give the potential V(x) as a function of x and the parameters of the disk.

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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Surface Charge Density
Surface charge density (σ) is defined as the amount of electric charge per unit area on a surface. In this case, the charge density varies with the distance from the center of the disk, given by the equation σ = aR², where 'a' is a constant and 'R' is the radial distance from the center. Understanding this concept is crucial for calculating the electric field and potential generated by the charged disk.
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Surface Charge Density
Electric Potential
Electric potential (V) at a point in space is the work done per unit charge in bringing a positive test charge from infinity to that point. It is a scalar quantity and can be influenced by the distribution of charge in the surrounding area. In this problem, we need to find the potential along the x-axis due to the nonuniform charge distribution on the disk.
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Integration in Electric Field Calculations
When dealing with continuous charge distributions, such as the nonuniform surface charge density in this problem, integration is often required to calculate the total electric field or potential. This involves summing contributions from infinitesimal charge elements across the entire surface. The integration process allows us to account for the varying charge density and its effect on the potential at different points along the x-axis.
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Finding Moment Of Inertia By Integrating
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