the charge to the left of p is 3Q and to the right is Q For

the charge to the left of p is +3Q and to the right is -Q.
For the charge distribution given above, each gridline represents 1 cm and Q= 2*10^-4C. a negative test charge q=-5*10^-5 C and m=12 *10^-7 kg is then placed at point P and released. Make a model (system schema, energy bar charts, electric potential lines and word explanation of the motion) and calculate the speed of the test charge when it is 0.1 cm away from the positive charge

Solution

initial potential energy = -k3Qq/3.62x10^-2 + kQq/1.414 x 10^-2 = -82.87kQq+70.72kQq=-12.15 kQq

final potential energy= -k3Qq/0.1 x 10^-2 +kQq/4 x 10^-2= -2975 kQq

change in potential energy = 2962.85 x kQq= 2962.85 x 9 x 10^9 x 2 x 10^-4 x 5 x 10^-5= 266656.62 j

1/2 x 12 x 10^-7 x v^2 = 266656.62

v= 66.66 x 10^4 m/s