Question

A proton is released from rest inside a region of constant, uniform electric field ?1 pointing due north. 34.8 s after it is released, the electric field instantaneously changes to a constant, uniform electric field ?2 pointing due south. 8.49 s after the field changes, the proton has returned to its starting point. What is the ratio of the magnitude of ?2 to the magnitude of ?1? You may neglect the effects of gravity on the proton.

Answer 1

We know that force on a positive charge particle due to electric field is given by:

Fe = q*E1

And direction of force is in the direction of electric field.

Using force balance:

F_net = Fe

m*a1 = q*E1

a1 = q*E1/m

Now Using 2nd kinematic equation, time taken (t1 = 34.8 sec) by proton to travel 'd' distance will be:

d = U*t1 + (1/2)*a1*t1^2

U = Initial speed of proton = 0 m/s, So

d = 0*t1 + (1/2)*(q*E1/m)*t1^2

d = q*E1*t1^2/(2m)

t1 = sqrt [2*m*d/(q*E1)]

Now after that electric field is instantaneously changed in opposite direction to E2, So at this time velocity of proton is:

V = U + a1*t = 0 + (q*E1/m)*t1

Now new acceleration of proton in opposite direction will be (Assuming north is +ve)

m*a2 = -q*E2

a2 = -q*E2/m

Now given that after time t2 (8.49 sec) protons returned at same starting point, So Using 2nd kinematic equation:

-d = V*t2 + (1/2)*a2*t2^2

Here -d, since electron returns at same point, So it displacement will be in negative and equal to 'd'

-q*E1*t1^2/(2m) = (q*E1/m)*t1*t2 + (1/2)*(-q*E2/m)*t2^2

multiply by m/q on both sides:

-E1*t1^2/2 = E1*t1*t2 - E2*t2^2/2

E2*t2^2/2 = E1*(t1*t2 + t1^2/2)

E2/E1 = (t1^2 + 2*t1*t2)/t2^2

Using given values of t1 and t2

E2/E1 = (34.8^2 + 2*8.49*34.8)/8.49^2 = 24.999

E2/E1 = 25.0

Let me know if you've any query.