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Chapter 20: Problem 31

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.

Short Answer

Expert verified
The magnitude of the electric field due to a charged ring of total charge \(Q\) and radius \(a\) on the ring axis at distance \(a\) from the ring's center is \( \frac{{k_eQ}}{{2a\sqrt{2}}} \).

Step by step solution

01

Define the Differential Element

Consider a very small charge element \(dQ\) of the ring. This element produces a small electric field \(dE\) at the point \(P\).
02

Calculate the Differential Electric Field

The magnitude of the differential electric field \(dE\) at point \(P\) due to \(dQ\) can be calculated using Coulomb’s law: \( dE = k_e \cdot \frac{{dQ}}{{r^2}} \), where \( k_e \) is Coulomb’s constant, \(r\) is the distance from \(dQ\) to \(P\), and in this case \(r = \sqrt{{a^2 + a^2}} = a\sqrt{2}\). Substituting the value of \(r\) in the equation, we get \( dE = k_e \cdot \frac{{dQ}}{{2a^2}} \).
03

Resolve the Differential Electric Field

Since the problem involves a ring, the electric field is symmetrical along the axis of the ring. The electric field's horizontal component due to \(dQ\) and the one due to the charge opposite to it on the ring will cancel each other, and only the vertical component will contribute to the total electric field. The vertical component of \(dE\) is \(dE\sin(\theta) = dE \cdot \frac{{a}}{{r}} = k_e \cdot \frac{{dQ}}{{2a^2}} \cdot \frac{{a}}{{a\sqrt{2}}} = k_e \cdot \frac{{dQ}}{{2a\sqrt{2}}}\).
04

Integrate Over Entire Ring

To get the total electric field at point \(P\), we sum up the electric fields from all such elements along the ring. This can be done via integration from 0 to \(Q\): \( E = \int dE\sin(\theta) = \int_0^Q k_e \cdot \frac{{dQ}}{{2a\sqrt{2}}} = \frac{{k_eQ}}{{2a\sqrt{2}}} \).

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