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Chapter 20: Problem 33

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 at a point on the axis of the ring, at a distance \(a\) from the center of the ring, is given by \(E = k\frac{Q}{2\sqrt{2}a^2}\), where \(k\) is approximately \(8.99 \times 10^9 N m^2/C^2\).

Step by step solution

01

Writing out the given values

First, let's write out the given values from the problem: \(Q\) is the total charge on the ring, \(a\) is the radius of the ring, and \(a\) is also the distance from the center of the ring to the point on the axis. Hence, we have \(r = a\) and \(x = a\).
02

Substituting into the formula for the electric field

Substitute the values from Step 1 into the formula for the electric field due to a charged ring: \(E = k\frac{Qx}{(x^2+r^2)^\frac{3}{2}}\), which leads to \(E = k\frac{Qa}{(a^2+a^2)^\frac{3}{2}}\). Simplify this expression further to get \(E = k\frac{Qa}{(2a^2)^\frac{3}{2}}\).
03

Simplifying the expression further

In this step, the expression will be simplified further. You'll find that \(E = k\frac{Q}{2\sqrt{2}a^2}\).
04

Using the value of Coulomb鈥檚 constant

The value of Coulomb鈥檚 constant \(k\) is approximately \(8.99 \times 10^9 N m^2/C^2\). Substitute this value into the expression to calculate the electric field.

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