A proton is released from rest inside a region of constant, uniform electric field E1 pointing due North. 14.8 seconds after it is released, the electric field instantaneously changes to a constant, uniform electric field E2 pointing due South. 8.07 seconds after the field changes, the proton has returned to its starting point. What is the ratio of the magnitude of E2 to the magnitude of E1? You may neglect the effects of gravity on the proton.
I found that the ratio is just a2/a1. I am not getting the right answer.
m = mass of proton
q = charge of proton
During first E1
F = q E1 north
a = q E1/m north
vfinal = (q/m) E1 (14.8)
x final = 0 + 0 +.5 (q/m)E1 (14.8)^2 = (q/m)E1 (110)
Now during E2
x initial = 110 (q/m)E1
Vinitial = (q/m)E1 (14.8)
x final = 0
so
0 = 110(q/m)E1 + 14.8(q/m)E1(8.07)-.5(q/m)E2(8.07^2)
so
.5 E2(8.07)^2 = 110 E1 + 14.8(8.07)E1
32.6 E2 = 110 E1+119 E1 = 229 E1
E2/E1 = 7.04
To solve this problem, we need to use the formula for the displacement of an object under constant acceleration:
s = ut + 0.5at^2
where s is the displacement, u is the initial velocity, a is the acceleration, and t is time.
Let's break down the problem into two separate cases: first when the proton is under the influence of electric field E1 and then when it is under the influence of electric field E2.
Case 1 (under E1):
In this case, the proton experiences a constant acceleration due to the electric field E1 pointing due North. Since it starts from rest, its initial velocity u is zero. The displacement s1 can be calculated using the formula above:
s1 = 0 + 0.5a1t1^2
where a1 is the acceleration due to E1 and t1 is the time it takes for the proton to return to its starting point.
Case 2 (under E2):
In this case, the electric field changes to E2 pointing due South. The proton experiences an acceleration due to E2 during this time. The displacement s2 can be calculated similarly:
s2 = 0 + 0.5a2t2^2
where a2 is the acceleration due to E2 and t2 is the time it takes for the proton to return to its starting point.
Given that s2 = s1 and the times are given, we can set up the following equation:
0 + 0.5a1(t1^2) = 0 + 0.5a2(t2^2)
Now, let's solve for the ratio of the magnitude of E2 to the magnitude of E1:
(a2/a1) = (t1/t2)^2
Substituting the given values of t1 and t2, we can calculate this ratio.
To find the ratio of the magnitude of E2 to the magnitude of E1, we can use the equation for displacement under constant acceleration:
$\Delta x = v_i t + \frac{1}{2} a t^2$
where $\Delta x$ is the displacement, $v_i$ is the initial velocity, $t$ is the time, and $a$ is the acceleration.
Let's break down the problem into two parts, before and after the field changes:
Before the field changes:
The proton is accelerated in the direction of the electric field E1, pointing due North. As the proton is released from rest, its initial velocity is 0.
Using the above equation, we can find the displacement of the proton before the field changes:
$\Delta x_1 = \frac{1}{2} a_1 t_1^2$
where $\Delta x_1$ is the displacement, $a_1$ is the acceleration due to E1, and $t_1$ is the time before the field changes.
After the field changes:
The proton is accelerated in the direction of the electric field E2, pointing due South. Since the proton returns to its starting point, its final displacement is 0.
Using the above equation, we can find the displacement of the proton after the field changes:
$\Delta x_2 = 0 = v_{i2} t_2 + \frac{1}{2} a_2 t_2^2$
where $\Delta x_2$ is the displacement, $v_{i2}$ is the initial velocity after the field changes, $a_2$ is the acceleration due to E2, and $t_2$ is the time after the field changes.
Now, let's solve the equations and find the values of $a_1$ and $a_2$:
From the equation $\Delta x_1 = \frac{1}{2} a_1 t_1^2$, we know that the displacement before the field changes is equal to 0 since the proton returns to its starting point. Therefore, we can write:
$0 = \frac{1}{2} a_1 t_1^2$
This leads to $a_1 = 0$ since $t_1 \neq 0$.
For the second part, solving the equation $\Delta x_2 = v_{i2} t_2 + \frac{1}{2} a_2 t_2^2$, we know that $v_{i2}$ is equal to the final velocity before the field changes. Hence, $v_{i2} = a_1 t_1$. Substituting this into the equation, we get:
$0 = a_1 t_1 t_2 + \frac{1}{2} a_2 t_2^2$
Since $a_1 = 0$, the equation simplifies to:
$0 = \frac{1}{2} a_2 t_2^2$
Solving for $a_2$, we find:
$a_2 = 0$
From the above results, we can see that both $a_1$ and $a_2$ are 0, implying that there is no acceleration and the electric fields do not affect the proton's motion.
Therefore, the ratio of the magnitude of E2 to the magnitude of E1 is also 0.