TheE-field strength is 50,000 N/C inside a parallel plate capacitor with 2.0 mmspacing. A proton is released from rest at the positive plate. What is the proton’sspeed when it reaches the negative plate?

The speed is

Explanation:

Electric field magnitude (E) and electric force magnitude (F) are related by the equation

with q the charge of the proton (). Because between parallel plates electric field is almost constant, electric force is constant too:

because electric force is constant, then by Newton's second law acceleration (a) is constant too, it is:

with m the mass of the proton ():

Now with this constant acceleration we can use the kinematic equation

with v the final speed, the initial velocity that is zero (because proton starts at rest) and d is the distance between the plates, so:

work: w = fd (given)

work done by chet   =50*1 = 50 j

work done by mika =40*2 = 80 j

work done by sara =30*1.5 = 45 j

work done by bill =60*0.5 = 30 j

students in order from the greatest amount of work done to the least is mika> chet> sara> bill

hence option (b) is correct.

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