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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

Point charges \(q_{1}=+2.00 \mu \mathrm{C}\) and \(q_{2}=-2.00 \mu \mathrm{C}\) are placed at adjacent corners of a square for which the length of each side is \(3.00 \mathrm{~cm}\). Point \(a\) is at the center of the square, and point \(b\) is at the empty corner closest to \(q_{2}\). Take the electric potential to be zero at a distance far from both charges. (a) What is the electric potential at point \(a\) due to \(q_{1}\) and \(q_{2} ?\) (b) What is the electric potential at point \(b ?\) (c) A point charge \(q_{3}=-5.00 \mu \mathrm{C}\) moves from point \(a\) to point \(b .\) How much work is done on \(q_{3}\) by the electric forces exerted by \(q_{1}\) and \(q_{2} ?\) Is this work positive or negative?

An object with charge \(q=-6.00 \times 10^{-9} \mathrm{C}\) is placed in a region of uniform electric field and is released from rest at point \(A .\) After the charge has moved to point \(B, 0.500 \mathrm{~m}\) to the right, it has kinetic energy \(3.00 \times 10^{-7} \mathrm{~J}\). (a) If the electric potential at point \(A\) is \(+30.0 \mathrm{~V}\). what is the clectric potential at point \(B ?\) (b) What are the magnitude and direction of the electric field?

A metal sphere with radius \(R_{1}\) has a charge \(Q_{1}\). Take the electric potential to be zero at an infinite distance from the sphere. (a) What are the electric field and electric potential at the surface of the sphere? This sphere is now connected by a long, thin conducting wire to another sphere of radius \(R_{2}\) that is several meters from the first sphere. Before the connection is made, this second sphere is uncharged. After electrostatic equilibrium has been reached, what are (b) the total charge on each sphere; (c) the electric potential at the surface of each sphere; (d) the clectric ficld at the surface of each sphere? Assume that the amount of charge on the wire is much less than the charge on each sphere.

A small sphere with mass \(5.00 \times 10^{-7} \mathrm{~kg}\) and charge \(+7.00 \mu \mathrm{C}\) is released from rest a distance of \(0.400 \mathrm{~m}\) above a large horizontal insulating sheet of charge that has uniform surface charge density \(\sigma=+8.00 \mathrm{pC} / \mathrm{m}^{2} .\) Using energy methods, calculate the speed of the sphere when it is \(0.100 \mathrm{~m}\) above the sheet.

A point charge \(q_{1}=+2.40 \mu \mathrm{C}\) is held stationary at the origin. A second point charge \(q_{2}=-4.30 \mu \mathrm{C}\) moves from the point \(x=0.150 \mathrm{~m}, y=0\) to the point \(x=0.250 \mathrm{~m}, y=0.250 \mathrm{~m} .\) How much work is done by the electric force on \(q_{2} ?\)
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