A negatively charged hockey puck is traveling along frictionless ice on a straig

Question

A negatively charged hockey puck is traveling along frictionless ice on a straight path into the center of a hockey goal. The defending team's goalie, lacking the skill to deflect the puck, instead puts equal negative charges on each side of the goal, 2.00 m apart, at floor level. The puck slows down on the way in, but the goalie did not use enough charge and the puck makes it into the goal anyway. (a) What is the net force on the puck when it is directly between the two charges, right on the goal line? (b) At what distance from the goal line did the puck experience the strongest negative acceleration? Express your answer as a positive number. A negatively charged hockey puck is traveling along frictionless ice on a straight path into the center of a hockey goal. The defending team's goalie, lacking the skill to deflect the puck, instead puts equal negative charges on each side of the goal, 2.00 m apart, at floor level. The puck slows down on the way in, but the goalie did not use enough charge and the puck makes it into the goal anyway. (a) What is the net force on the puck when it is directly between the two charges, right on the goal line? (b) At what distance from the goal line did the puck experience the strongest negative acceleration? Express your answer as a positive number.

Explanation / Answer

charge on the puck -Q and the charges at the goal -q. the two - q charges are separated by L = 2 m.

Lets take the puck as moving along the x axis so the two charges -q are located at (0, -L/2) and (0, L/2)

for the puck at a distance x from the center of the goal, we can write the forces due to the two chares -q. Define a as the angle made by a line connecting the puck with either of the -q charges and the x axis. The we can write:

Force in x direction: Fx = k(-Q)(-q)/(x^2 +(L/2)^2) cos(a) + k(-Q)(-q)/(x^2 + (-L/2)^2) cos(a)

but cos(a) = x/sqrt(x^2+(L/2)^2) so

Fx = 2kQq x/(x^2 + (L/2)^2)^(3/2)

You can see by symmetry that Fy = 0 since the charges each puysh the puck in opposite directions in y with the same magnitude force.

Now let's answer part a. Set x = 0 and

Fx = 2kQq x/(x^2 + (L/2)^2)^(3/2) = 0 which it has to be and Fy = 0 from above

For part b Use F = ma so

ma = 2kQq x/(x^2 + (L/2)^2)^(3/2) Now to find max a, take da/dx, set = 0 and solve for x:

da/dx = 0 = 2kQq/m*{1/(x^2 +(L/2)^2)^(3/2) - 3x^2/(x^2 +(L/2)^2)^(5/2)}

0 = 1 - 3x/(x^2 + (L/2)^2) ---> x^2 + (L/2)^2 = 3x ---> x^2 - 3x + (L/2)^2 =0

Use quadratic formula

x = 3/2 +/- 1/2 sqrt(9 - (L/2)^2) = 3/2 +/- 1/2 sqrt(8) = 3/2 +/- 2*sqrt(2)

Choose the "-" in teh calculation so that x = 3/2 - 2*sqrt(2) = -1.38 m or x = 1.38 m

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