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

$(9.0\times {10}^{9}N{m}^2/C^2)\frac{(2.0\times {10}^{-9}C)s}{{(0.025m)}^3}=1150N/C$

Short answer

(a) A dipole with charge $\pm q(2\times {10}^{-9}C)$and a distance $s$between them.

At a distance of $0.025m$from the dipole, the electric field is $1150N/C$.

In a dipole, what is the charge separation?

(b) The solution is$\mathbf{1}\mathbf{}\mathit{m}\mathit{m}$

Step-by-step solution

Given information and formula used

Given :

$(9.0\times {10}^{9}N{m}^2/C^2)\frac{(2.0\times {10}^{-9}C)s}{{(0.025m)}^3}=1150N/C$

Theory used :

Anelectric dipole deals with the separation of the positive and negative charges found in any electromagnetic system.

Writing a realistic problem and finding the solution of the problem

(a) Realistic problem :

A dipole with charge $\pm q(2\times {10}^{-9}C)$and a distance $s$between them.

At a distance of $0.025m$from the dipole, the electric field is $1150N/C$.

In a dipole, what is the charge separation?

(b) We can safely assume that $r>>s$.

$$\begin{gathered} 1150N/C=(9.0\times {10}^{9}N{m}^2/C^2)\frac{(2.0\times {10}^{-9}C)s}{{(0.025m)}^3} \\ \Rightarrow s=\frac{(1150N/C)({(0.025m)}^3}{(9.0\times {10}^{9}N{m}^2/C^2)(2.0\times {10}^{-9}C)} \\ =1\times {10}^{-3}m \\ =\boxed{\mathbf{1}\mathbf{}\mathit{m}\mathit{m}} \end{gathered}$$