Question

A proton is released from rest inside a region of constant, uniform electric field E 1 pointing due north. 27.3 s after it is released, the electric field instantaneously changes to a constant, uniform electric field E 2 pointing due south. 3.03 s after the field changes, the proton has returned to its starting point. What is the ratio of the magnitude of E 2 to the magnitude of E 1 ? You may neglect the effects of gravity on the proton.

Answer 1

Answer:

E2/E1 =99.2

Explanation:

time after release of E1 (t) = 27.3 s

time after release of E2 (t') = 3.03 s

acceleration (a) = $\frac{QE1}{M}$

where

• Q is the protons charge
• M is the mass

after 27.3 s

velocity (V) = a x t = $\frac{QE1}{M}$ x 27.3  = $\frac{27.3QE1}{M}$

distance to turning point (s) = 0.5a$t^{2}$ = 0.5 x $\frac{QE1}{M}$x $27.3^{2}$ = $\frac{372.65QE1}{M}$

now for its return back to its starting point

acceleration (a') = $-\frac{QE2}{M}$

total distance S' = distance to turning point + distance from turning point to starting point

S' = S + vt' + 0.5 a'$t'^{2}$

S' = $\frac{372.65QE1}{M}$ + ($\frac{27.3QE1}{M}$ x 3.03) + (0.5 x $-\frac{QE2}{M}$x $3.03^{2}$)

S' is the distance at the starting point and = 0

0 = $\frac{372.65QE1}{M} + \frac{82.72QE1}{M}-\frac{4.59QE2}{M}$

$\frac{4.59QE2}{M}=\frac{372.65QE1}{M} + \frac{82.72QE1}{M}$

multiplying both side by M/Q we have

4.59.E2 = 372.65E1 + 82.72E1

4.59.E2 = 455.37E1

E2/E1 = 455.37 / 4.59

E2/E1 =99.2