Question

At time t=0 a proton is a distance of 0.360 m from a very large insulating sheet of charge and is moving parallel to the sheet with speed 960 m/s . The sheet has uniform surface charge density 2.34.×10−9C/m2 . What is the speed of the proton at 5.40

Answer 1

Explanation:

Formula for the electric field due to the infinite sheet of charge is as follows.

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

where,   $\sigma$ = surface charge density

Now, formula for electric force acting on the proton is as follows.

             F = eE

where,    e = charge of the proton

According to the Newton's second law of motion, the net force acting on the proton is as follows.

                       F = ma

                 a = $\frac{eE}{m}$

                    = $\frac{e(\frac{\sigma}{2 \epsilon_{o}})}{m}$

                    = $\frac{e \sigma}{2m \epsilon_{o}}$

According to the kinematic equation, speed of the proton in perpendicular direction is as follows.

              $v_{f} = v_{i} + at$

                     = $(0 m/s) + \frac{e \sigma}{2 m \epsilon_{o}}t$

                     = $\frac{1.6 \times 10^{-19}C \times 2.34 \times 10^{-9} C/m^{2} \times 5.40 \times 10^{-8}s}{2 \times (1.67 \times 10^{-27} kg)(8.85 \times 10^{-12} C^{2}/Nm^{2}}$

                     = 683.974 m/s

Hence, total speed of the proton is as follows.

                v' = $\sqrt{(960 m/s)^{2} + (683.974 m/s)^{2}}$

                    = $\sqrt{921600 + 467820.43}$

                    = $\sqrt{1389420.43}$

                    = 1178.73 m/s

Therefore, we can conclude that speed of the proton is 1178.73 m/s.