Question

In Problems 63 through 66 you are given the equation(s) used to solve a problem. For each of these

a. Write a realistic problem for which this is the correct equation(s).

b. Finish the solution of the problem

$\frac{\eta }{2{\epsilon }_0}[1-\frac{z}{\sqrt{z^2+R^2}}]=\frac{1}{2}\frac{\eta }{2{\epsilon }_0}$

Short answer

(a) A disc with a radius of $R$and a charge of $Q$that is uniformly distributed. Discover an equation for the strength of the electric field at a point $P$on the axis at a distance $z$from the center.

(b) The solution is$\mathit{z}\mathbf{=}\frac{\mathbf{R}}{\sqrt{\mathbf{3}}}$

Step-by-step solution

Given information and formula used

Given :

$\frac{\eta }{2{\epsilon }_0}[1-\frac{z}{\sqrt{z^2+R^2}}]=\frac{1}{2}\frac{\eta }{2{\epsilon }_0}$

Theory used :

The electric field due to a uniformly charged disc at a point very distant from the surface of the disc is given by:

$E=\frac{\sigma }{2{\epsilon }_0}$

($\sigma$ is the surface charge density on the disc)

Writing a realistic problem and finding the solution of the problem 

(a) Realistic problem :

Given a disc with a radius of $R$and a charge of $Q$that is uniformly distributed. Discover an equation for the strength of the electric field at a point $P$on the axis at a distance $z$from the center. If

$\frac{\eta }{2{\epsilon }_0}[1-\frac{z}{\sqrt{z^2+R^2}}]=\frac{1}{2}\frac{\eta }{2{\epsilon }_0}$

(b) Solution :

From the previous expression, finding the value of z :

$$\begin{gathered} \frac{\eta }{2{\epsilon }_0}[1-\frac{z}{\sqrt{z^2+R^2}}]=\frac{1}{2}\frac{\eta }{2{\epsilon }_0} \\ \Rightarrow 1-\frac{z}{\sqrt{z^2+R^2}}=\frac{1}{2} \\ \Rightarrow \frac{z^2}{z^2+R^2}=\frac{1}{4} \\ \Rightarrow z^2+R^2=4z^2 \\ \Rightarrow \boxed{\mathit{z}\mathbf{=}\frac{\mathbf{R}}{\sqrt{\mathbf{3}}}} \end{gathered}$$