Q37P - Fundamentals Of Physics

Question

Suppose you design an apparatus in which a uniformly charged disk of radius R is to produce an electric field. The field magnitude is most important along the central perpendicular axis of the disk, at a point P, at distance  $\mathbf{2}\mathbf{.00}\mathit{R}$from the disk (Fig. 22-57a). Cost analysis suggests that you switch to a ring of the same outer radius R but with inner radius $\mathit{R}\mathbf{/}\mathbf{2}\mathbf{.00}$ (Fig. 22-57b). Assume that the ring will have the same surface charge density as the original disk. If you switch to the ring, by what percentage will you decrease the electric field magnitude at P?

Step-by-Step Solution

Answer

If we switch to the ring, the percentage decrease of the electric field magnitude at P is.$28\%$

 

Step 1: The given data

• 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 radius and inner radius $R/2$ now is to produce a field.

 

Step 2: Understanding the concept of electric 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,                            $E=\frac{\sigma }{2{\epsilon }_o}(1-\frac{z}{\sqrt{z^2+R^2}})$                                                        (i)

where,  = surface charge density

z = distance on the central axis of the disk  

R = Radius of the disk

Step 3: Calculation of the decreased percentage

 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:

$E_{\left(b\right)}=E_{\left(a\right)}-\frac{\sigma }{2{\epsilon }_o}\left(1-\frac{2R}{\sqrt{{\left(2R\right)}^2+{\left(\frac{R}{2}\right)}^2}}\right)$

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

$E_{\left(a\right)}=\frac{\sigma }{2{\epsilon }_o}\left(1-\frac{2R}{\sqrt{{\left(2R\right)}^2+{\left(R\right)}^2}}\right)$

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

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

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