91Ó°ÊÓ

• A uniformly charged disk of radius,R is to produce an electric field at a distance 2R of a point from the central perpendicular axis of the disk.
• A ring of the same outer radiusand inner radius$R/2$now is to produce a field.

Using the concept of the electric field at a point by comparing their given data in their respective electric fields formulae, we can get the required decrease percentage in the electric field of the second case.

Formula:

The magnitude of the electric field produced by the disk at a point on its central axis, $E=\frac{\mathrm{σ}}{2{\mathrm{Î\mu }}_o}(1−\frac{z}{\sqrt{z^2+R^2}})$ (i)

where,= surface charge density

z = distanceon the central axis of the disk

R = Radius of the disk

We can see that the disk in Figure (b) is effectively equivalent to the disk in Figure (a) plus a concentric smaller disk (of radius$R/2$) with the opposite value of.

That electric field by the ring is given using equation (i) as:

role="math" localid="1661915871465" $E_{\left(b\right)}=E_{\left(a\right)}−\frac{\mathrm{σ}}{2{\mathrm{Î\mu }}_o}\left(1−\frac{2R}{\sqrt{{\left(2R\right)}^2+{\left(\frac{R}{2}\right)}^2}}\right)$

And the electric of the disk is given using same equation (i) as:

$E_{\left(a\right)}=\frac{\mathrm{σ}}{2{\mathrm{Î\mu }}_o}\left(1−\frac{2R}{\sqrt{{\left(2R\right)}^2+{\left(R\right)}^2}}\right)$

Thus, the percentage decrease in the electric field can be given using the above two equations as:

$\begin{array}{c}decreased\%=\frac{E_{\left(a\right)}−E_{\left(b\right)}}{E_{\left(a\right)}}×100\%\\ =\frac{1−2/\sqrt{4+1/4}}{1−2/\sqrt{4+1}}×100\%\\ =\frac{0.0299}{0.1056}×100\%\\ =28.3\%\\ ≈28\%\end{array}$

Hence, the value of the decreased percentage is.$28\%$