03. A proton is placed at point A between the two oppositely-charged parallel pl
ID: 1792765 • Letter: 0
Question
03. A proton is placed at point A between the two oppositely-charged parallel plates Problem 03 shown in the diagram to the right. The potential difference between the plates is 3000 V and the plates are 16.0 cm apart. a.) What is the electrostatic force (magnitude) acting on this charge at point A? b.) What is the potential difference between points A and B? c.) If the charge is released from A and allowed to accelerate to point B, what will be its velocity when it reaches point B? d.) What is the minimum work done in moving the charge from point A to C?Explanation / Answer
q=+1.6*10^-19 C, V=3000V, d=0.16m
a)
V=E*d
E=V/d =3000/0.16 = 18750 V/m
F=qE = (1.6*10^-19)*(18750) = 3.0*10^-15 N
b)
V= E*x = (18750)(0.09) = 1687.5 V
c)
Use work-Energy theorem,
W=qV
1/2mv^2 = qV
v=sqrt[2qV/m]
v=sqrt[(2*1.6*10^-19*1687.5)/(1.67*10^-27)] = 5.7*10^5 m/s
d)
WA to B = q*V= (1.6*10^-19)*(1687.5) = 2.7*10^-16 J
WB to C = q*V= (1.6*10^-19)*(0) = 0 J
WA to C = WA to B + WB to C = 2.7*10^-16 J + 0 J = 2.7*10^-16 J
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