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)
V1 + V2 = 500

k*q1/r + k*q2/(8-r) = 500

9*10^9*4*10^-6/r - 9*10^9*6*10^-6/(8-r) = 500

36*10^3*(8-r-r) = 500*(r*(8-r))

72*(8-2r) = 8r - r^2

576 - 144r = 8r - r^2

r^2 - 152r + 576 = 0

solving above eqn we

get r = 3.89 m

so at x = -5 + 3.89 = -1.11 m V = 500 volts

potentail at origin.

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

here r1 = 5 m, r2 = 3m

V = -1.08*10^4 volts

potential at x = 2m,

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

here r1 = 7 m, r2 = 1m

V' = -4.886*10^4 volts

Potential difference, = v-v' = 3.806*10^4 volts

electrons move towards -x direction


b)

U1 = k*q1*q2/r1

here r1 = 8m

U1 = -0.027 J

U2 = k*q1*q2/r2

here r1 = 0.2m

U2 = -1.08 J


Workdone = U1-U2 = 1.053 J

2*0.5*m*v^2 = 1.053

v = sqrt(1.053/m)

if m = 1 kg, v = 1.026 m/s

if m = 1 gm , v = 32.45 m/s

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