The electric field strength is 5.50×104 N/C inside a parallel-plate capacitor with a 2.30 mm spacing. A proton is released from rest at the positive plate. What is the proton's speed when it reaches the negative plate?

v = 1.09*10^5 m/s

Explanation:

In order to calculate the speed of the proton when it reaches the negative plate, you first calculate the acceleration of the proton, produced by the electric force.

You use the second Newton law:

           (1)

q: charge of the proton = 1.6*10^-19 C

m: mass of the proton = 1.67*10^-27 kg

E: magnitude of the electric field between the plates = 5.50*10^4N/C

You solve for the acceleration a in the equation (1):

Next, you use the following formula to find the final speed of the proton:

          (2)

v: final speed of the proton = ?

vo: initial speed of the proton = 0m/s

x: distance traveled by the proton = 2.30mm/2 = 1.15mm = 1.15*10^-3m

You replace the values of all parameters in the equation (2):

The speed of the proton when it reaches the negative plate is 1.09*10^5 m/s

answer; equivalent to a one(1) horsepower motor;

note: the given question is incomplete.

complete question is given below.

which is an example of kinetic energy?

a. the energy stored in ethanol  

b. a hockey puck sliding across the ice  

c. a compressed spring  

d. a ball sitting at the top of a ramp

solution: the kinetic energy is the energy posses by the virtue of the motion of the object.

in the given options, a, c and d are the example of potential energy because it is the energy posses by the virtue of the rest of the object. but option b has the kinetic energy as the object is in motion.

hence, a hockey puck sliding across the ice   is the apex example of kinetic energy.

so, the option (b) a hockey puck sliding across the ice   is the correct option.