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Chapter 20: Problem 4

Why should the test charge used to measure an electric field be small?

Short Answer

Expert verified
A small test charge should be used to measure an electric field to ensure an accurate measurement, as a larger test charge can distort the field by introducing its own electric field.

Step by step solution

01

Understand Electric Field Calculation

An electric field is calculated using Coulomb's Law, given as \(E = F/q\), where \(F\) is the force experienced by a charge, and \(q\) is the charge itself. This formula tells us how much force is felt by the charge due to an electric field.
02

Reason for Small Test Charge

If a large test charge is used, it can distort the electric field being measured because it introduces a significant amount of charge and creates its own electric field. This can interfere with the original electric field that is being measured. This is why a small test charge is used - to minimize the impact on the electric field and increase the accuracy of the measurement.
03

Verifying the Requirement of a Small Test Charge

The purpose of a test charge is to test the field, not affect it. If a large charge was used, it would create an interaction making compensation necessary and complicating the measurement. Therefore, a small test charge is used as it ensures that the reading of the electric field is accurate, as it would feel the field but not significantly distort it.

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Most popular questions from this chapter

Find the magnitude of the electric force on a \(2.0-\mu \mathrm{C}\) charge in a \(100-\mathrm{N} / \mathrm{C}\) electric field.

A thin rod lies on the \(x\) -axis between \(x=0\) and \(x=L\) and carries total charge \(Q\) distributed uniformly over its length. Show that the electric field strength for \(x>L\) is given by \(E=k Q /[x(x-L)]\)

Find the magnitude of the electric field due to a charged ring of radius \(a\) and total charge \(Q\) on the ring axis at distance \(a\) from the ring's center.

A straight wire 10 m long carries \(25 \mu C\) distributed uniformly over its length. (a) What's the line charge density on the wire? Find the electric field strength (b) \(15 \mathrm{cm}\) from the wire axis, not near either end, and (c) 350 m from the wire. Make suitable approximations in both cases.

Find the line charge density on a long wire if the electric field \(45 \mathrm{cm}\) from the wire has magnitude \(260 \mathrm{kN} / \mathrm{C}\) and points toward the wire.
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