Question

The electric potential at a position located a distance of 22.5 mm from a positive point charge of 8.20×10^-9C and 13.0 mm from a second point charge is 1.06 kV. Calculate the value of the second charge.

Also...

A positive charge of 4.40 μC is fixed in place. From a distance of 4.40 cm a particle of mass 6.00 g and charge +3.90 μC is fired with an initial speed of 66.0 m/s directly toward the fixed charge. How close to the fixed charge does the particle get before it comes to rest and starts traveling away?

Answer 1

q1 = 8.2*10^-9 C

distance of q1 from point P = r1 = 22.5 mm = 22.5*10^-3 m

distance of q2 from poin p = r2 = 13 mm = 13*10^-3 m

potantial due to q1 = v1 = k*q1/r1

potential due to q2 = V2 = k*q2/r2

total potential V = v1 + v2

given V = 1.06*10^3 v

1.03*10^3 = k*q1/r1 + k*q2/r2

1030 = (9*10^9)*((8.2*10^-9)/(22.5*10^-3) + q2/(13*10^-3) )

q2 = -3.25*10^-9 C <<<<<<----------answer

++++++++++++++

initial kinetic energy = KE1 = 0.5*m*v1^2

initial potential energy = k*q1*q2/r1

total initial energy = TE1 = KE1 + PE1

TE1 = 0.5*m*v1^2 + k*q1*q2/r1

after a distance of distance r2 the comes to rest

final kinetic energy = KE2 = 0

final potential energy = PE2 = k*q1*q2/r2

total final energy TE2 = KE2 + PE2 = k*q1*q2/r2

from energy conservation

TE2 = TE1

k*q1*q2/r2 = 0.5*m*v1^2 + k*q1*q2/r1

k*q1*Q2*( (1/r2) - (1/r1) ) = 0.5*m*v1^2

(9*10^9*4.4*10^-6*3.9*10^-6)*( 1/r2) - (1/0.044) ) = 0.5*6*10^-3*66^2

r2 = 0.93 cm   <<<------------answer