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

Knight Calc 5th Edition

Ch 25: The Electric Potential

Problem 72

Chapter 25, Problem 72

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.

Verified step by step guidance

Step 1: Recall the formula for electric potential due to a point charge, which is given by \( V = \frac{kQ}{r} \), where \( k \) is Coulomb's constant, \( Q \) is the charge, and \( r \) is the distance from the charge to the point of interest.

Step 2: Recognize that the semicircular rod is composed of many infinitesimal charge elements \( dq \), each contributing to the electric potential at the center. Since electric potential is a scalar quantity, the contributions from all charge elements can be summed directly.

Step 3: Express the total charge \( Q \) distributed uniformly along the semicircular rod. The linear charge density \( \lambda \) is defined as \( \lambda = \frac{Q}{L} \), where \( L \) is the length of the semicircle. For a semicircle of radius \( R \), \( L = \pi R \). Thus, \( \lambda = \frac{Q}{\pi R} \).

Step 4: Write the infinitesimal charge element \( dq \) in terms of \( \lambda \) and the infinitesimal arc length \( ds \): \( dq = \lambda ds \). Since \( ds \) is the arc length of the semicircle, \( ds = R d\theta \), where \( \theta \) is the angle subtended by the arc at the center.

Step 5: Integrate the contributions to the electric potential from all charge elements along the semicircle. The distance from each charge element to the center is constant and equal to \( R \). The total potential is \( V = \int \frac{k dq}{R} = \int \frac{k \lambda ds}{R} = \int \frac{k \lambda R d\theta}{R} \). Simplify and evaluate the integral over \( \theta \) from \( 0 \) to \( \pi \) to find the expression for the electric potential at the center.

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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 that indicates the work done in bringing a unit positive charge from infinity to that point. The electric potential due to a continuous charge distribution can be calculated by integrating the contributions from each infinitesimal charge element.

Charge Distribution

Charge distribution refers to how electric charge is spread over a given region. In this case, the charge Q is uniformly distributed along the semicircular rod. Understanding the nature of the charge distribution is crucial for calculating the electric potential, as it affects how the electric field is generated and how it influences points in space, particularly at the center of the semicircle.

Integration in Physics

Integration is a mathematical tool used in physics to calculate quantities that accumulate over a continuous distribution, such as charge or mass. In the context of finding electric potential, integration allows us to sum the contributions of infinitesimal charge elements along the semicircular rod to find the total potential at the center. This process is essential for dealing with non-point charge distributions.

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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

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 P25.67 shows two uniformly charged spheres. What is the potential difference between points 1 and 2? Which point is at the higher potential? Hint: The potential at any point is the superposition of the potentials due to all charges.

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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

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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