/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 43 A wire is bent in the form of a ... [FREE SOLUTION] | 91Ó°ÊÓ

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A wire is bent in the form of a regular hexagon and a total charge \(q\) is distributed uniformly on it. What is the electric field at the centre? You may answer this part without making any numerical calculations.

Short Answer

Expert verified
The net electric field at the center is zero due to symmetry.

Step by step solution

01

Understand the Problem

The problem involves a regular hexagon formed by a charged wire with a total charge \( q \) uniformly distributed. Our goal is to determine the electric field at the center of the hexagon.
02

Consider Symmetry

A regular hexagon has a high degree of symmetry, with each of its sides equally spaced from the center. When charge is uniformly distributed along these sides, the resultant electric field at the center due to each segment will interact symmetrically.
03

Determine Electric Field Contribution from a Side

Due to symmetry, the contribution of the electric field from one side will be balanced by the field from its opposite side. This is because the electric field created by a line of charge at a point perpendicular to its center is directed along the perpendicular, and opposite sides will have electric fields in opposite directions.
04

Calculate Net Electric Field

Since opposite sides of the hexagon cancel out each other's electric field at the center due to symmetry, the net electric field at the center is zero. This happens because for each pair of sides that are directly opposite, their contributions to the electric field at the center cancel each other out completely.

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Key Concepts

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

Charged Wire
Imagine a wire that carries an electric charge and is bent into the shape of a geometric figure. In our case, it's a regular hexagon. The wire has a charge, denoted as \( q \), that is evenly distributed along its length.

A charged wire creates an electric field around it. The electric field is due to the fact that each tiny piece of the wire, since it carries charge, emits an electric field. When these charged sections are lined up to form a hexagon, the fields interact in a structured way. Understanding these interactions is crucial for determining the effect at different points, particularly at the hexagon's center in this exercise.

Considering a problem involving a charged wire requires analyzing how the charge distribution affects the space around it, specifically focusing on symmetry and field cancellation in this case.
Symmetry in Electric Fields
A regular hexagon exhibits symmetry, meaning each segment or side of the hexagon is identical in shape and charge distribution. This symmetry is a powerful tool in physics because it allows us to make predictions about how electric fields will behave without needing complex mathematics.

When a charge is evenly spread across the hexagon, the electric field due to each side has unique interactions because of this symmetry. In simpler terms, at the center of the hexagon, each side produces an electric field that points in a direction perpendicular to the midpoint of the side. Due to the symmetry of the hexagon, opposites sides have fields that directly oppose each other.

This principle of symmetry simplifies calculations; rather than calculating each field individually and adding them up, we can predict that effects from symmetrical components will cancel each other, leading us to conclude that the electric field at the center is zero.
Cancellation of Electric Fields
The cancellation of electric fields in symmetrical arrangements is a fascinating phenomenon. Because of the hexagon's shape and the uniform distribution of charge along its sides, each side's electric field contribution cancels out that of its opposite side.

Here's why this happens:
  • Each pair of opposite sides in a regular hexagon generates electric fields that are equal in magnitude but opposite in direction at the center.
  • This opposition causes a complete cancellation of their effects due to the vector nature of electric fields. Vectors pointing in opposite directions sum to zero.
As a result, no matter how strong or weak the fields from individual sides are, their net effect at the exact center is zero.

This is a great example of how symmetry and proper understanding of vector addition in electric fields can lead to elegant and simple solutions in physics.

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Most popular questions from this chapter

A particle of mass \(m\) and charge \(q\) is thrown at a speed \(u\) against a uniform electric field \(E .\) How much distance will it travel before coming to momentary rest?

An electric field of \(20 \mathrm{~N} \mathrm{C}^{-1}\) exists along the \(x\) -axis in space. Calculate the potential difference \(V_{B}-V_{A}\) where the points \(A\) and \(B\) are given by, (a) \(A=(0,0) ; B=(4 \mathrm{~m}, 2 \mathrm{~m})\) (b) \(A=(4 \mathrm{~m}, 2 \mathrm{~m}) ; B=(6 \mathrm{~m}, 5 \mathrm{~m})\) (c) \(A=(0,0) ; B=(6 \mathrm{~m}, 5 \mathrm{~m})\) Do you find any relation between the answers of parts (a), (b) and (c) ?

Three equal charges, \(2.0 \times 10^{-6} \mathrm{C}\) each, are held fixed at the three corners of an equilateral triangle of side \(5 \mathrm{~cm}\). Find the Coulomb force experienced by one of the charges due to the rest two.

A particle of mass \(1 \mathrm{~g}\) and charge \(2.5 \times 10^{-4} \mathrm{C}\) is released from rest in an electric field of \(1.2 \times 10^{4} \mathrm{~N} \mathrm{C}^{-1}\). (a) Find the electric force and the force of gravity acting on this particle. Can one of these forces be neglected in comparison with the other for approximate analysis ? (b) How long will it take for the particle to travel a distance of \(40 \mathrm{~cm} ?\) (c) What will be the speed of the particle after travelling this distance ? (d) How much is the work done by the electric force on the particle during this period?

Two identically charged particles are fastened to the two ends of a spring of spring constant \(100 \mathrm{~N} \mathrm{~m}^{-1}\) and natural length \(10 \mathrm{~cm}\). The system rests on a smooth horizontal table. If the charge on each particle is \(2 \cdot 0 \times 10^{-8} \mathrm{C}\), find the extension in the length of the spring. Assume that the extension is small as compared to the natural length. Justify this assumption after you solve the problem.

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