A group of sleep deprived engineering student just invented yet another new game
ID: 2990069 • Letter: A
Question
A group of sleep deprived engineering student just invented yet another new game call "Electrostatic Darts. The game consists of a square dart board 2 m on each side. There is a charge placed on each of the 4 corners. A player stands 2 m from the board and throws a dart also charged with Q. Find the total energy needed to move the dart from the initial position 2 m from the board to the center of the board. If 2 opposing corners were now replaced with -Q what is the total energy needed to move the dart from the initial position 2 m from the board to the center of the board. [Concept = very little work] In the original configuration (part a) would it be likely (or possible) to hit one of the 4 corner charges? Explain.Explanation / Answer
Half of a diagonal of the board (distance from one corner to the center) is
R1=D/2 =L*sqrt(2)/2 =sqrt(2)=1.41 m
Potential at the board center from 4 equal charges Q is
V1 =4*KQ/R1 = (2.828*KQ)
that is energy of the dart being positioned in the board center is W1 =Q*V1 =4KQ^2/R1
Distance fron each charge to where is located initial the dart before being thrown (2 m from board center)
R2 =sqrt(R1^2 + 2^2) =sqrt(2+4) =sqrt(6)
Potential at the player location is V2 =4KQ/R2
Energy of the dart at the player location W2 =Q*V2 =4KQ^2/R2
Energy necessary to move the dart from player to center of board is
W(final-initial)=W12 =W1-W2 =4KQ^2*(1/R1 -1/R2) =4KQ^2*(1/sqrt(2) -1/sqrt(6)) =1.195*KQ^2 >0
Work need to be done for the dart to arrive at the center of the board.
b)
If 2 diagonal oposimg charges are replaced with -Q then the total potential at the center of the board is 0.
V1 =2KQ/R1 -2KQ/R1 =0
Final energy (dart in center of board) W1 =Q*V1 =0
Potential at player location (providede player is positioned symetrically from all 4 charges) is again zero for the same reason.
V2 =0
Initial energy before throwing W2 -Q*V2 =0
Energy nedded to move dart from player to board center is
W12 =W1-W2 =0 -0=0 J
c)
Potential at one corner location is
V1' =2KQ/L +KQ/(L*sqrt(2)) =KQ +KQ/2.828 =1.353*KQ
Assuming player position remains unchanged:
W'(final -initial) =W12' =W1' -W2= 1.353KQ^2 -4KQ^2/sqrt(6) =-0.279*KQ^2 <0 <W12
It is easier to hit one corner of the table than to hit its center.
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