charge q1=-2.90x10^-5 C is at position (0,1.20 m) and charge q2=1.40x10^-5 C is
ID: 1329592 • Letter: C
Question
charge q1=-2.90x10^-5 C is at position (0,1.20 m) and charge q2=1.40x10^-5 C is at position (-0.500 m,0).
a) what is the potential at postion p (0.500 m,0)?
b) if an electron is placed at p what would be its potential energy?
c) if the electron is released from point p with zero velocity calculate its speed when it crosses am equipotential surface which includes the origin.
d) how much work would have to be done by an external agent so that the electron travels from point p to that equipotential surface with constant speed?
Explanation / Answer
x1 , y1 = 0 , 1.2
x2,y2 = -0.5 , 0
x3, y3 = 0.5 , 0
r1 = sqrt(x3-x1)^2+(y3-y1)^2 = sqrt((0.5-0)^2+(0-1.2)^2) = 1.3 m
r2 = sqrt(x3-x2)^2+(y3-y2)^2 = sqrt((0.5+0.5)^2+(0-0)^2) = 1.0 m
a)
V1 = k*q1/r
v2 = k*q2/r2
Vp = v1 + v2 = k*q1/r1 + k*q2/r2
vp = -(9*10^9*2.9*10^-5)/1.3 + (9*10^9*1.4*10^-5)/1.0 = -74769.23 v <<------answer
b)
Up = vp*e = 74769.23*1.6*10^-19 = 1.19*10^-14 J <<------answer
part(c)
potentail at origin
vo = k*q1/y1 + k*q2/x2
vo = -(9*10^9*2.9*10^-5)/1.2 + (9*10^9*1.4*10^-5)/0.5 = 34500 v
Uo = voe = 34500*1.6*10^-19 = 5.52*10^-15 J
change in PE = change in KE
Up-Uo = 0.5*m*v^2
(1.19*10^-14) - (5.52*10^-15) =0.5*9.11*10^-31*v^2
v = 1.18*10^8 m/s <<------answer
d)
W = -(Uo-Up) = -6.38*10^-15 <<------answer
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