1) Two charges q1 = + 3.00 muC and q2 = -4.00muC are located at positions (-2,0)
ID: 1397504 • Letter: 1
Question
1) Two charges q1 = + 3.00 muC and q2 = -4.00muC are located at positions (-2,0) and (3,0)on the X - axis as shown in the figure. The distance are measured in cm = 001m. Calculate (using K = 9.00 x 10^9): A) The Coulomb force between the charges, and if it is attractive or repulsive? B) The field of each charge, E1 and E2 at the point A(0,0), indicating with arrows their directions (E or W) C) The net field there, and its direction (E, W) D) The potential produced by each charge at A (0,0) and the net value VA there. E) Calculate also the potentials of each charge at point B (0,4) and their net value VB there. F) The work done by a test charge of +2.00 muC when moving from A to B G) Finally, the potential energy U12 of the two charges q1 and q2Explanation / Answer
a) The Coulomb force between two opposite charges is always attractive.
b) E = kq/r^2
for q1, E= 9 x 10^9 ( 3 x 10^-6)/ (0.02)^2= 6.75 x 10^7 ( pointing towards E)
for q2, E = 9 x 10^9 ( -4 x 10^-6) / (0.03)^2 =- 4 x 10^7 ( pointing towards E)
c) Net electric field = (6.75-4)x 10^7 = 2.75 x 10^7 ( towards E)
d) V = kq/r
due to q1= (9 x 10^9 )( 3 x 10^-6)/ (0.02) = 1350 x 10^3
due to q2 = ( 9 x 10^9 )(-4 x 10^-6)/( 0.03) = -1200 x 10^3
net VA=( 150 x 10^3)
e) V at B
due to q1= (9 x 10^9 )( 3 x 10^-6)/ 0.045 = 600 x 10^3 apprx
due to q2 = (9 x 10^9 )(-4 x 10^-6)/ 0.05 = - 720 x 10^3
net VB= (-120 x 10^3)
f) work done= 2 x 10^-6( -120 - 150) x 10^3 = -540 x 10^-3 J
g) g) Potential energy = 9x 10^9 ( 3 x 10^-6)( -4 x 10^-6)/ (0.05) = 2160 x 10^-3
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