A uniform electric field exists in the region between two oppositely charged plane parallel plates. A proton is released from rest at the surface of the positively charged plate and strikes the surface of the opposite plate, 1.60 cm distant from the first, in a time interval of 3.20×10−6s3.20×10 −6 s. (a) Find the magnitude of the electric field. (b) Find the speed of the proton when it strikes the negatively charged plate.

E = 326.17 N/C

Explanation:

(a) In order to calculate the magnitude of the electric field between the parallel plates you first calculate the acceleration of the proton. You use the following formula:

        (1)

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

t: time that the proton takes to cross the space between the plates = 3.20*10^-6 s

a: acceleration of the proton = ?

x: distance traveled by the proton = 1.60cm = 0.016m

You solve the equation (1) for a, and replace the values of all parameters:

Next, you use the Newton second law for the electric force, to find the magnitude of the electric field:

          (2)

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

m: mass of the proton = 1.77*10^-27kg

You solve the equation (2) for E:

The magnitude of the electric field in between the parallel plates is 326.17N/C

what is the speed of a softball that travels 20m in 2secs

speed = 20/2 = 10 m/s

you can cut your hand off if your not