“Electrostatic Darts”

A group of sleep deprived engineering students 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 Q placed on each of the 4 corners. A player stands 2 m from the board and throws a dart also charged with Q

a. Find the total energy needed to move the dart from the initial position 2 m from the board to the center of the board.

b. 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.

c. [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. [Hint: would it take more or less energy than hitting the center?] 

a)

Half of a diagonal of the board (distance from one corner to the center) is

$R_1=D/2 =L*sqrt{2}/2 =sqrt{2}=1.41 m$

Potential at the board center from 4 equal charges Q is

$V_1 =4*K Q/R_1 = (2.828*K Q)$

that is energy of the dart being positioned in the board center is

$W_1 =Q*V_1 =4KQ^2/R_1$

Distance from each charge to where is located initial the dart before being thrown (2 m from board center)

$R_2 =sqrt{R_1^2 + 2^2} =sqrt{2+4} =sqrt{6}$

Potential at the player location is $V2 =4KQ/R_2$

Energy of the dart at the player location $W_2 =Q*V_2 =4KQ^2/R_2$

Energy necessary to move the dart from player to center of board is

$W(final-initial)=W_{12} =W_1-W_2 =4KQ^2*(1/R_1 -1/R_2) =$

$=4KQ^2*(1/sqrt{2} -1/sqrt{6}) =1.195*K Q^2 > 0$

Work need to be done for the dart to arrive at the center of the board.

b)

If 2 diagonal opposite charges are replaced with $-Q$ then the total potential at the center of the board is 0.

$V_1 =2KQ/R_1 -2KQ/R_1 =0$

Final energy (dart in center of board) $W_1 =Q*V_1 =0$

Potential at player location (provided player is positioned symmetrically from all 4 charges) is again zero for the same reason.

$V_2 =0$

Initial energy before throwing $W2 =-Q*V2 =0$

Energy needed to move dart from player to board center is

$W_{12} =W_1-W_2 =0 -0=0 J$

c)

Potential at one corner location is

$V_1′ =2KQ/L +K Q/(L*sqrt{2}) =K Q +K Q/2.828 =1.353*K Q$

Assuming player position remains unchanged:

$W'(final -initial) =W_{12}’ =W_1′ -W_2= 1.353*K Q^2 -4KQ^2/sqrt{6} =$

$=-0.279*K Q^2 <0 <W_{12}$

It is easier to hit one corner of the table than to hit its center.