Answer:
The voltage drop through which the proton moves is 39.1 V.
Step-by-step explanation:
Given that,
Distance = 4.76 cm
Time

We need to calculate the acceleration
Using equation of motion

Where, s = distance
a = acceleration
t = time
Put the value in the equation



We need to calculate the voltage drop
Using formula of electric field

....(I)
Using newton's second law
....(II)
Put the value of F in equation (I) from equation (II)


Where, q = charge
a = acceleration
d = distance
m= mass of proton
Put the value into the formula


Hence, The voltage drop through which the proton moves is 39.1 V.