Question

A proton moving to the right at 6.2 × 10^5 ms^-1 enters a region where there is an electric field of 62 kNC^-1 directed to the left. Describe qualitatively the motion of the proton in this filed. What is the time taken by the proton to come back to the point where it entered the field? (Use the standard values of the mass and charge of a proton) Be as descriptive as possible, please!

Answer 1

Given

   speed of protons is v =  6.2× 10⁵ ms^-1

   Electric field E =  62kNC^-1 = 62 *10³NC^-1

   charge q = 1.6*10^-19C

   Mass of the proton is m=1.67*10^-27kg

We have the formula for force is given by

F = q E

Apply Newtons second law

F= m a

so that   

a = q E /m

=(1.60x10^-19)(62,000) / (1.67x10^-27)= - 5.94x10¹² m/s²

We have put a minus sign on this value of "a" because it is adeceleration (ie, the force is to the left and the motion is to theright so that the proton is slowing down).

Now, v_f = v_o + a t where the finalvelocity is zero when it stops over to the right somewhere.

0= 6.2x10⁵ - 5.94x10¹²t,     solve for t to get 1.044x10^-7second.

This time is HALF of the total time since the proton turnsaround and then accelerates to the left with the same value of "a", just positive this time.

Thus, the total time to return to starting point is2.088x10^-7 sec.