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

$(9.0\times {10}^{9}N{m}^2/C^2)\frac{2(2.0\times {10}^{-7}C/m)}{r}=25000N/C$

Short answer

(a) Find the Electric field strength at point $P$on the rod's axis at distance $r$from the center of an infinite charge rod, with the linear charge density is$2\times {10}^{-7}C/m$in $25000N/C$

(b) The solution is$\mathbf{0}\mathbf{.}\mathbf{144}\mathbf{}\mathit{m}$

Step-by-step solution

Given information and formula used

Given :

$(9.0\times {10}^{9}N{m}^2/C^2)\frac{2(2.0\times {10}^{-7}C/m)}{r}=25000N/C$

Theory used :

TheElectric field of an infinite line charge with a uniform linear charge density can be obtained by a using Gauss' law. Considering a Gaussian surface in the form of a cylinder at radius $r$, the electric field has the same magnitude at every point of the cylinder and is directed outward.

Writing a realistic problem and finding the solution of the problem 

(a) Realistic problem :

On an infinite charge rod, let the linear charge density is $2\times {10}^{-7}C/m$.

In $25000N/C$, find the electric field strength at point $P$on the rod's axis at distance $r$from the center.

(b) Solution :

$$\begin{gathered} (9.0\times {10}^{9}N{m}^2/C^2)\frac{2(2.0\times {10}^{-7}C/m)}{r}=25000N/C \\ \Rightarrow r=(9.0\times {10}^{9}N{m}^2/C^2)\frac{2(2.0\times {10}^{-7}C/m)}{25000N/C} \\ =\boxed{\mathbf{0}\mathbf{.}\mathbf{144}\mathbf{}\mathit{m}} \end{gathered}$$