Ch. 23 - Electric Potential

Giancoli Douglas - Physics for Scientists and Engineers 5th edition

Giancoli Douglas5th editionPhysics for Scientists and EngineersISBN: 9780137488179Not the one you use?Change textbook

Giancoli Douglas 5th edition

Ch. 23 - Electric Potential

Problem 38

Chapter 22, Problem 38

A thin rod of length 2ℓ is centered on the x axis as shown in Fig. 23–46. The rod carries a uniformly distributed charge Q. Determine the potential V as a function of y for points along the y axis. Let V = 0 at infinity.

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Start by recalling the formula for the electric potential due to a point charge: \( V = \frac{k_e q}{r} \), where \( k_e \) is Coulomb's constant, \( q \) is the charge, and \( r \) is the distance from the charge to the point of interest. For a continuous charge distribution, the potential is calculated by integrating over the charge distribution: \( V = \int \frac{k_e dq}{r} \).

Set up the coordinate system. The rod is centered on the x-axis, with its length extending from \( -\ell \) to \( \ell \). The point of interest is along the y-axis at a distance \( y \) from the origin. The distance \( r \) from a point on the rod to the point on the y-axis is given by \( r = \sqrt{x^2 + y^2} \), where \( x \) is the position along the rod.

Express the charge element \( dq \) in terms of the linear charge density \( \lambda \), where \( \lambda = \frac{Q}{2\ell} \). The charge element is then \( dq = \lambda dx \), where \( dx \) is an infinitesimal segment of the rod.

Substitute \( dq \) and \( r \) into the integral for the potential: \( V = \int_{-\ell}^{\ell} \frac{k_e \lambda dx}{\sqrt{x^2 + y^2}} \). This integral accounts for the contribution to the potential from each infinitesimal segment of the rod.

Solve the integral. Since \( \lambda \) and \( k_e \) are constants, they can be factored out of the integral: \( V = k_e \lambda \int_{-\ell}^{\ell} \frac{dx}{\sqrt{x^2 + y^2}} \). Use a standard integral formula for \( \int \frac{dx}{\sqrt{x^2 + a^2}} \), which evaluates to \( \ln(x + \sqrt{x^2 + a^2}) \). Apply the limits of integration from \( -\ell \) to \( \ell \) to find the final expression for \( V \).

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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 (V) is the amount of electric potential energy per unit charge at a point in an electric field. It is a scalar quantity and is measured in volts (V). The potential due to a charged object can be calculated by integrating the contributions from each infinitesimal charge element, taking into account the distance from the charge to the point of interest.

Coulomb's Law

Coulomb's Law describes the force between two point charges. It states that the force (F) between two charges is directly proportional to the product of the magnitudes of the charges (Q1 and Q2) and inversely proportional to the square of the distance (r) between them. This law is fundamental in calculating the electric field and potential due to distributed charges.

Superposition Principle

The superposition principle states that the total electric potential at a point due to multiple charges is the algebraic sum of the potentials due to each charge considered separately. This principle allows for the calculation of the potential from a continuous charge distribution, such as the uniformly charged rod in the question, by integrating the contributions from each infinitesimal segment of the rod.

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