Question

The surface charge density on an infinite charged plane is - 2.10 ×10−6C/m2. A proton is shot straight away from the plane at 2.40 ×106m/s. How far does the proton travel before reaching its turning point?

Answer 1

Explanation:

Formula to calculate electric field because of the plate is as follows.

         E = $\frac{\sigma}{2 \times \epsilon_{o}}$

            = $\frac{2.10 \times 10^{-6}}{2 \times 8.85 \times 10^{-12}}$

           = $1.18 \times 10^{5} N/C$

Now, we will consider that equilibrium of forces are present there. So,

                   ma = qE

       a = $\frac{1.6 \times 10^{-19} \times 1.18 \times 10^{5}}{1.67 \times 10^{-27}}$

          = $1.13 \times 10^{13} m/s^2$

According to the third equation of motion,

         $v^{2} = 2 \times a \times d$

or,      d = $\frac{v^{2}}{2d}$

             = $\frac{(2.4 \times 10^{6})^{2}}{2 \times 1.13 \times 10^{13}}$

             = 0.254 m

Thus, we can conclude that the proton will travel 0.254 m before reaching its turning point.