Question

A proton orbits a long charged wire, making $1.0\times {10}^6$ revolutions per second. The radius of the orbit is $1.0cm$. What is the wire’s linear charge density?

Short answer

The wire’s linear charge density is$\mathbf{-}\mathbf{2}\mathbf{.}\mathbf{28}\mathbf{}\mathit{n}\mathit{C}\mathbf{/}\mathit{m}$

Step-by-step solution

Given information and Formula used

Given :

Proton makes revolutions per second : $1.0\times {10}^6$

The radius of the orbit : $1.0cm$

Theory used :

Electric field due to long charged wire at distance $r$

$E=\frac{2\lambda }{4\pi {\epsilon }_{0}r}$

(Electrostatic force is attractive, therefore $\lambda$is negative.)

The proton receives centripetal force from electrostatic force, that is$F_c=F_e$

Calculating the wire’s linear charge density 

Now, $PE=m{w}^{2}r$

Using the above relation, we can deduce :

$$\begin{gathered} \frac{-2\lambda \times e}{2\pi {\epsilon }_{0}r}=m{w}^{2}r \\ \Rightarrow -\lambda =\frac{1.67\times {10}^{-27}\times {(2\mathrm{\pi }\times {10}^6)}^2\times ({10}^{-2}{)}^2}{2\times 1.6\times {10}^{-19}\times 9\times {10}^9} \\ =-2.28x{10}^{9}C/m \\ \Rightarrow \lambda =\boxed{\mathbf{-}\mathbf{2}\mathbf{.}\mathbf{28}\mathbf{}\mathit{n}\mathit{C}\mathbf{/}\mathit{m}} \end{gathered}$$