The electric potential at a position located a distance of 22.5 mm from a positi
ID: 1398673 • Letter: T
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 C is fixed in place. From a distance of 4.40 cm a particle of mass 6.00 g and charge +3.90 C 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?
Explanation / Answer
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
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