A charge of +.4microC is at (-5, 0) meters and a charge of -.6microC is at (+3,

Question

                    A charge of +.4microC is at (-5, 0) meters and a charge of -.6microC is at (+3, 0) meters.
                

                    a) Where is it +500 volts on the x-axis? If there is a 2 meter rod from (0, 0) to (0, 2) meters, what is the potential difference between the two ends of the                     rod?Which direction will the electrons in the rod flow?                 

                    b) If the charges are allowed to come together freely, where will they end up & what is their final velocity?                 

Explanation / Answer

a)V = k*q1/x - k*q2/(d-x)


V *x*(d-x)= k *[ d*q1-x*q1-xq2 ]

dx - x^2 = k/V *[ d*q1-x*q1-xq2 ]


8x-x^2 = 57.6 - 18x

x^2 -26x+57.6 = 0

x =23.55 m or x = 2.44m

potentail at origin.

V = V1+V2 = k*Q1/r1+k*Q2/r2

here r1 = 5 m, r2 = 3m....q1 = 0.6*10^-6C.....q2 = 0.4*10^-6C

V = (9*10^9*0.6*10^-6/5) - (9*10^9*0.4*10^-6/5) = 360 volts

potential at x = 2m,

V' = V1+V2 = k*Q1/r1+k*Q2/r2

here r1 = sqrt(3^2+2^2) = sqrt( 9+4) = sqrt13 =3.61 m, r2 = sqrt5^2+2^2) = sqrt(29) = 5.34m

V' = (9*10^9*0.6*10^-6/3.61) - (9*10^9*0.4*10^-6/5.34) = 821.68 volts

Potential difference, = v' - v = 461.68 volts


v <v'

electrons move towards low potential to high potential   +y direction
b) U1 = k*q1*q2/r1

here r1 = 8m

U1 = 0.00027 J

r2 = 0.2m


U2 = k*q1*q2/r2

here
U2 = 0.0108J


Workdone = U1-U2 = 0.01053J

2*0.5*m*v^2 = 0.01053

v = sqrt(0.01053/m)

if m = 1 m kg = 0.001kg, v = 3.24 m/s

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