Question

The surface charge density on an infinite charged plane is $-2.0\times {10}^{-6}C/m^2$ . A proton is shot straight away from the plane at $2.0\times {10}^{6}m/s$. How far does the proton travel before reaching its turning point?

Short answer

The proton travels$\mathbf{18}\mathbf{}\mathit{c}\mathit{m}$ before reaching its turning point .

Step-by-step solution

Given information and formula used

Given :

The surface charge density on the infinite charged plane : $-2.0\times {10}^{-6}C/m^2$

Proton shot straight away at the speed : $2.0\times {10}^{6}m/s$

Theory used :

The Electric field owing to an infinite plane is calculated as follows:

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

Newton's law :

$\frac{1}{2}m({\left(v_f\right)}^2–{\left(v_i\right)}^2)=qEd$

Calculating how far the proton travels before reaching its turning point 

Due to the action of the infinite plane's electric field, this proton will come to a halt before returning. We can compute the distance it goes before turning back using conservation of energy :

$$\begin{gathered} \frac{1}{2}m({\left(v_f\right)}^2–{\left(v_i\right)}^2)=qEd \\ \Rightarrow d=\frac{m({\left(v_f\right)}^2–{\left(v_i\right)}^2)}{2qE} \\ =\frac{m({\left(v_f\right)}^2–{\left(v_i\right)}^2)}{2q(\sigma /2{\epsilon }_0)} \\ =\frac{(1.67x{10}^{-27}kg)(0-{(2.0x{10}^{6}m/s)}^2)(8.854x{10}^{-12}{C}^2/N·m^2)}{\left(1.6x{10}^{-19}C\right)(2.0x{10}^{-6}C/m^2)} \\ \Rightarrow d=\boxed{\mathbf{18}\mathbf{}\mathit{c}\mathit{m}} \end{gathered}$$