Refer to thr figure, the distance d shown is 0.0500 m. Point a is on the perpendicular bisector of the line connecting the two protons. Also, point a is a distance h = 12.00 cm = 0.1200 m above the line connecting the two protons. The proton charge is e = 1.6x10-19 C. a) A negative charge is released from rest at point a. (The other two protons remain at rest.) What is the speed v of this third proton when it is very far (i.e., at an infinite distance) from the other two protons? The electron mass is 9.11x10-31kg. b) What direction does the negative charge move? Go down or up? Why?
Refer to thr figure, the distance d shown is 0.0500 m. Point a is on the perpendicular bisector of the line connecting the two protons. Also, point a is a distance h = 12.00 cm = 0.1200 m above the line connecting the two protons. The proton charge is e = 1.6x10-19 C.
a) A negative charge is released from rest at point a. (The other two protons remain at rest.) What is the speed v of this third proton when it is very far (i.e., at an infinite distance) from the other two protons? The electron mass is 9.11x10-31kg.
b) What direction does the negative charge move? Go down or up? Why?
Given Data :
d = 0.0500 m
h = 12.00 cm = 0.1200 m
e = 1.6x10-19 C
electron mass m= 9.11x10-31kg
To Find :
speed v and direction
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