Question

A proton is initially at rest. After some time, a uniform electric field is turned on and the proton accelerates. The magnitude of the electric field is 1.36 105 N/C.
(a) What is the speed of the proton after it has traveled 3.00 cm?
(b) What is the speed of the proton after it has traveled 30.0 cm?

Answer 1

Answer:

a) 8.83*10⁵ m/s  b) 2.80*10⁶ m/s

Explanation:

a) Assuming no other forces acting on the proton, the acceleration on it is produced by the electric field.

By definition, the  force due to the electric field is as follows:

F = q*E = e*E (1)

where e is the elementary charge, the charge carried by only one proton, and is e = 1.6*10⁻¹⁹ C.

According to Newton's 2nd law, this force is at the same time, the product of the mass of the proton, times the acceleration a:

F = mp*a (2)

From (1) and(2), being left sides equal, right sides must be equal too:

$F = e*E = mp*a$

Solving for a:

$a = \frac{e*E}{mp} =\frac{1.6e-19C*1.36e5N/C}{1.67e-27kg} =1.3e13 m/s2$

⇒ a = 1.3*10¹³ m/s²

As we have the value of a (which is constant due to the field is uniform), the displacement x, and we know that the initial velocity is 0, in order to get the value of the speed, we can use the following kinematic equation:

$vf^{2} -vo^{2} = 2*a*x$

Replacing by v₀ = 0, a= 1.3*10¹³ m/s² and  x = 0.03 m, we can find vf as follows:

$vf =\sqrt{2*(1.3e13 m/s2)*0.03m} = 8.83e5 m/s$

⇒ vf = 8.83*10⁵ m/s

b) We can just repeat the equation from above, replacing x=0.03 m by x=0.3 m, as follows:

$vf =\sqrt{2*(1.3e13 m/s2)*0.3m} = 2.80e6 m/s$

⇒ vf = 2.80*10⁶ m/s