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Ch 25: The Electric Potential
Knight Calc - Physics for Scientists and Engineers 5th Edition
Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796Not the one you use?Change textbook
All textbooksKnight Calc 5th EditionCh 25: The Electric PotentialProblem 74
Chapter 25, Problem 74

The wire in FIGURE P25.74 has linear charge density λ. What is the electric potential at the center of the semicircle?

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Step 1: Understand the problem. The wire consists of a semicircular arc and two straight segments. The linear charge density of the wire is λ, and we need to calculate the electric potential at the center of the semicircle (denoted as the black dot in the figure). The electric potential is a scalar quantity, and contributions from different parts of the wire can be added directly.
Step 2: Recall the formula for electric potential due to a charged element. The potential at a point due to a small charge element dq is given by V = k_e * dq / r, where k_e is Coulomb's constant, dq is the charge of the element, and r is the distance from the charge element to the point of interest.
Step 3: Analyze the semicircular arc. The arc has radius R, and every point on the arc is equidistant from the center. The total charge on the arc is λ * (πR), where πR is the length of the semicircular arc. The potential contribution from the arc is V_arc = k_e * (λ * πR) / R.
Step 4: Analyze the straight segments. Each straight segment has a length of 2R. The distance from any point on the straight segment to the center varies, so we need to integrate to find the potential contribution. For a small element dx on the straight segment, dq = λ * dx, and the distance to the center is √(x^2 + R^2). The potential contribution from one segment is V_segment = ∫(k_e * λ * dx / √(x^2 + R^2)) from x = -2R to x = 0 for the left segment, and similarly for the right segment from x = 0 to x = 2R.
Step 5: Combine the contributions. The total electric potential at the center is the sum of the contributions from the semicircular arc and the two straight segments: V_total = V_arc + 2 * V_segment. Perform the integration for the straight segments and add the results to the potential from the arc to find the final expression for the electric potential.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Electric Potential

Electric potential, often denoted as V, is the amount of electric potential energy per unit charge at a point in an electric field. It is a scalar quantity measured in volts (V) and indicates the work done to move a unit positive charge from a reference point to a specific point in the field without any acceleration.
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Linear Charge Density

Linear charge density, represented by the symbol λ (lambda), is defined as the amount of electric charge per unit length along a line. It is expressed in coulombs per meter (C/m) and is crucial for calculating the electric field and potential generated by charged objects, such as the wire in the semicircular configuration.
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Integration in Electric Potential Calculation

To find the electric potential due to a continuous charge distribution, integration is often used. This involves summing the contributions to the potential from infinitesimal charge elements along the charged object. In this case, the semicircular wire's potential at the center requires integrating the contributions from each segment of the wire, taking into account the geometry and charge density.
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Related Practice
Textbook Question

An electric dipole consists of 1.0 g spheres charged to ±2.0 nC at the ends of a 10-cm-long massless rod. The dipole rotates on a frictionless pivot at its center. The dipole is held perpendicular to a uniform electric field with field strength 1000 V/m, then released. What is the dipole's angular velocity at the instant it is aligned with the electric field?

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Textbook Question

Two metal objects that are in contact must be at the same potential, an assertion we'll prove in the next chapter. Suppose a metal sphere of radius R is charged to 1000 V and a second metal sphere of radius 2R is charged to 2000 V. The two spheres are brought into contact and then separated. Afterward, what is the potential of each sphere?

Textbook Question

FIGURE P25.72 shows a thin rod with charge Q that has been bent into a semicircle of radius R. Find an expression for the electric potential at the center.

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Textbook Question

You are given the equation(s) used to solve a problem. Finish the solution of the problem: (9.0×109Nm2/C2)q₁q₂/0.030m =90×10−6J; q₁+q₂=40nC.

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Textbook Question

FIGURE P25.70 shows a thin rod of length L and charge Q. Find an expression for the electric potential a distance x away from the center of the rod on the axis of the rod.

Textbook Question

Bead A has a mass of 15 g and a charge of −5.0 nC. Bead B has a mass of 25 g and a charge of −10.0 nC. The beads are held 12 cm apart (measured between their centers) and released. What maximum speed is achieved by each bead?

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