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Chapter 23: Problem 68

A thin insulating rod is bent into a semicircular arc of radius \(a,\) and a total electric charge \(Q\) is distributed uniformly along the rod. Calculate the potential at the center of curvature of the are if the potential is assumed to be zero at infinity.

Short Answer

Expert verified
The electric potential at the center of curvature of the semicircular arc is \(k * Q/a\), where k is Coulomb's constant, Q is the total charge and a is the radius of the semi-circle.

Step by step solution

01

Find charge density

First, the linear charge density (\(\lambda\)) of the rod is found by dividing the total charge by the length of the rod. The length of a semi-circle is \(\pi*a\), therefore, the charge density is \(\lambda = Q / (\pi * a)\).
02

Defining an infinitesimal element

Next, determine a small charge (\(dq\)) on the rod which is at a distance \(r = a\) from the center of curvature. This distance remains constant over the semicircle due to the rod's shape. An infinitesimal angle \(\theta\) is described at the center by this charge. The length of this infinitesimal element will be \(a*d\theta\) and the charge \(dq\) would be \(\lambda*a*d\theta\).
03

Determining the electric potential

The electric potential (\(dV\)) due to infinitesimal charge \(dq\) is given by Coulomb's law as, \(dV = k*dq / r\). Substitute \(dq\) with \(\lambda*a*d\theta\) and \(r\) with \(a\) and this gives: \(dV = k* \lambda * d\theta\).
04

Integrating over the semi-circle

Now, add up these small contributions from each infinitesimal element over the whole semicircle through integration. The total potential at the center would be: \(V = \int_{0}^{\pi} dV = \int_{0}^{\pi} k*\lambda *d\theta\). After integrating, this results in: \(V = k * \lambda * \pi\).
05

Substitute the value of \(lambda\)

Finally, substitute the value of \(\lambda\) in terms of Q and a obtained in Step 1: \(V = k * Q / a\). Therefore, the electric potential at the center of curvature of the semicircular arc is \(k * Q/a\).

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Key Concepts

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

Semicircular Arc
Imagine a rod that is not straight but bent into a half-circle, akin to a smile. This shape is called a semicircular arc. A semicircular arc, as the name implies, is half of a full circular arc. The curvature is essential because it defines the specific path along which a charged object is distributed.
  • The radius of this semicircle is denoted as \( a \).
  • The half-circle covers an angular range from 0 to \( \pi \) radians.
  • With a complete circle having 360 degrees or \( 2\pi \) radians, the semicircle covers exactly half of a circle.
The semicircular configuration helps retain uniformity across the object, which simplifies calculating potential, especially when dealing with symmetrically distributed charges, such as in this exercise.
Charge Density
Charge density is an essential concept when dealing with distributed charges like on a semicircular arc. It tells us how much charge is present along the length of the rod. When we talk about linear charge density, it simply means the charge per unit length.
  • In this scenario, the total charge \( Q \) is distributed uniformly along the semicircular arc.
  • The mathematical representation of charge density \( \lambda \) is \( \lambda = \frac{Q}{\pi a} \).
  • This formula shows that charge density depends on the total charge and the length of the arc (which is the semicircle's circumference).
Understanding charge density allows for easy calculation of electric potential by giving a simple way to express how charge is distributed along a path like the semicircular arc.
Coulomb's Law
Coulomb's Law is a fundamental principle used to calculate the electric forces exerted by charges. It's crucial for determining the electric potential created by small charge elements:
  • The electric potential \( dV \) due to a small charge \( dq \) is described by the formula \( dV = \frac{k \cdot dq}{r} \).
  • Here, \( k \) is Coulomb's constant, representing the strength of the electric force between point charges.
  • \( r \), the distance from the charge to the point of interest, is always \( a \) because we're evaluating potential at the center of curvature, equidistant to all points on the semicircle.
Coulomb's Law allows us to integrate charge contributions over the semicircle via an organized mathematical approach, leading to the total potential without needing a large number of piecemeal calculations.
Infinitesimal Element
An infinitesimal element concept in calculus is of enormous importance in physics, particularly in dealing with continuous charge distributions. It refers to a tiny portion of a system we analyze.
  • On the semicircular arc, this infinitesimal element is a tiny part of the rod described by a small angle \( d\theta \).
  • The small charge \( dq \) at this segment is given by \( \lambda \cdot a \cdot d\theta \).
  • To find the total electric potential, we compute the contribution from each infinitesimal element and sum them up through integration over the arc from 0 to \( \pi \).
By considering these infinitesimal elements, we break down complex physical problems into manageable parts, ultimately simplifying calculations and ensuring our results reflect the continuous nature of charge distribution along the semicircular arc.

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Most popular questions from this chapter

At a certain distance from a point charge, the potential and electric-field magnitude due to that charge are \(4.98 \mathrm{~V}\) and \(16.2 \mathrm{~V} / \mathrm{m}\) respectively. (Take \(V=0\) at infinity.) (a) What is the distance to the point charge? (b) What is the magnitude of the charge? (c) Is the electric field directed toward or away from the noint charge?

A very small sphere with positive charge \(q=+8.00 \mu \mathrm{C}\) is released from rest at a point \(1.50 \mathrm{~cm}\) from a very long line of uniform linear charge density \(\lambda=+3.00 \mu \mathrm{C} / \mathrm{m} .\) What is the kinetic energy of the sphere when it is \(4.50 \mathrm{~cm}\) from the line of charge if the only force on it is the force exerted by the line of charge?

A heart cell can be modeled as a cylindrical shell that is \(100 \mu \mathrm{m}\) long, with an outer diameter of \(20.0 \mu \mathrm{m}\) and a cell wall thickness of \(1.00 \mu \mathrm{m}\), as shown in Fig. \(\mathrm{P} 23.81 .\) Potassium ions move across the cell wall, depositing positive charge on the outer surface and leaving a net negative charge on the inner surface. During the so-called resting phase, the inside of the cell has a potential that is \(90.0 \mathrm{mV}\) lower than the potential on its outer surface. (a) If the net charge of the cell is zero, what is the magnitude of the total charge on either cell wall membrane? Ignore edge effects and treat the cell as a very long cylinder. (b) What is the magnitude of the electric field just inside the cell wall? (c) In a subsequent depolarization event, sodium ions move through channels in the cell wall, so that the inner membrane becomes positively charged. At the cnd of this event, the inside of the cell has a potential that is \(20.0 \mathrm{mV}\) higher than the potential outside the cell. If we model this event by charge moving from the outer membrane to the inner membrane, what magnitude of charge moves across the cell wall during this event? (d) If this were done entirely by the motion of sodium ions, \(\mathrm{Na}^{+}\), how many ions have moved?

Two large, parallel conducting plates carrying opposite charges of equal magnitude are separated by \(2.20 \mathrm{~cm}\). (a) If the surface charge density for each plate has magnitude \(47.0 \mathrm{nC} / \mathrm{m}^{2}\), what is the magnitude of \(\dot{E}\) in the region between the plates? (b) What is the potential difference between the two plates? (c) If the separation between the plates is doubled while the surface charge density is kept constant at the value in part (a), what happens to the magnitude of the electric field and to the potential difference?

A uniformly charged, thin ring has radius \(15.0 \mathrm{~cm}\) and total charge \(+24.0 \mathrm{nC}\). An electron is placed on the ring's axis a distance \(30.0 \mathrm{~cm}\) from the center of the ring and is constrained to stay on the axis of the ring. The electron is then released from rest. (a) Describe the subsequent motion of the electron. (b) Find the speed of the electron when it reaches the center of the ring.
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