The three charges are held in place in the figure below, where L = 1.25 m. Find

Question

The three charges are held in place in the figure below, where L = 1.25 m. Find the electric potential at point P, at infinity, and the magnitude of their difference. Suppose that a fourth charge, with a charge of 6.65 mu C and a mass of 4.71-10^-3 kg, is released from rest at point P. What is the speed of the fourth charge when it has moved infinitely far away from the other three charges? Suppose your strange friend Bob insists his equation for the electric potential of a point charge is better: v_Bob = kQ/r + C where C = Bob's constant, based on Bob's favorite number = 61416 V What would be Bob's answer for the above questions? for part a: for part b: (Which answers changed? Which are important changes? Is Bob's formula better or worse than yours?)

Explanation / Answer

a) V = k*q/r = 9.0x109*(2.75x10-6/(1.25/2) + -1.72x10-6/(1.25/2) + 7.45x10-6/(1.25*sin(60))

=162000kV

b) U = V*q

= 16200V*6.65x10-6

   =0.1077 j

So at infinity K = 0.1077J

since K = 1/2*m*v^2

then v = sqrt(2*0.563/m) = sqrt(2*0.1077/4.71x10^-3) = 6.7m/s

(c)Vbob=KQ/r + C

C=61416

for (a)

162000kV+61416

=223416kV

for(b) 223416V*6.65x10-6

=1.48 j

So at infinity K = 1.48J

since K = 1/2*m*v^2

then v = sqrt(2*0.563/m) = sqrt(2*1.48/4.71x10^-3) =25.05m/s

There may be calculation error as it include many calculation , but concept is right , i have tried to solve upto best of my knowledge

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