Q#2 What is the electrical potential at the origin due to the two 2.40 A mu C ch

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Q#2

What is the electrical potential at the origin due to the two 2.40 A mu C charges? Two particles, with charges of 40.0 nc and -40.0 nC, are placed at the points with coordinates (0, 8.00) cm and (0, -8.00) cm as shown in Figure P2S.28. A particle with charge 20.0 nC is located at the origin. Find the electric potential energy of the configuration of the three fixed charges. A fourth particle, with a mass of 1.95 x 10^13 kg and a charge of 80.0 nC, is released from rest at the point (6.00, 0) cm. Find its speed after it has moved freely to a very large distance away. Calculate the energy required to assemble the array of charges shown in Figure P2S.33, where a = 0.200 m, b = 0.400 m, and q = 1.00 mu C. Assume a reference level of potential V = 0 at r = infinity.

Explanation / Answer

Given charges are at

   q1 = 4 nC at (0,8)cm

   q2 = -4 nC, at (0,-8)cm

and a particle with charge q3 = 20 nC at (0,0) cm

now the potential energy , of two charges separated by a distance d is

   U1,2 = k*q1*q2/r12

here

   U = U12+U13+U23

       U12 = k*q1*q2/r12
       = 9*10^9*4*(-4)*10^-18/(16*10^-2) J

       = -9*10^-7 J

   U13 = k*q1*q3/r13
   = 9*10^9*4*(20)*10^-18/(8*10^-2) J
   = 9*10^-6 J

U23 = k*q2*q3/r23
= 9*10^9*(-4)*(20)*10^-18/(8*10^-2) J

= -9*10^-6 J

total potential energy of the system is

   U = (-9*10^-7 )+(9*10^-6)+(-9*10^-6) J

   U = -9*10^-7 J

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