A point charge q 2 = 0.2 ?C is fixed at the origin of a co-ordinate system as sh
ID: 252013 • Letter: A
Question
A point charge q2 = 0.2 ?C is fixed at the origin of a co-ordinate system as shown. Another point charge q1 = 4.6 ?C is is initially located at point P, a distance d1 = 5 cm from the origin along the x-axis
A point char 20.2 HC is fixed at the origin of a co-ordinate system as shown. Another point charge 1 4.6 HC is is initially located at point P, a distance d1 5 cm from the origin along thex 1) What is PE, the change in potenial energy of charge q 1 when it is moved from point p to point R, located a distance d2 2 cm from the origin along the x-axis as shown? J Submit 2) The charge q2 is now replaced by two charges q3 and q4 which each have a magnitude of 0.1 LC, half of that of q2. The charges are located a distance a = 1.2 cm from the origin along the y-axis as shown. What is pE, the change in potential energy now if charge q1 is moved from point P to point R? teiae 12 cm from ho oragitutle d o.1 3) q3 What is the potential energy of the system composed of the three charges q1, q3, and q4, when g1 is at point R? Define the potential energy to be zero at infinity.Explanation / Answer
(1) Change in potential energy = 8.98*109 *4.6*10-6*0.2*10-6/0.02 - 8.98*109 *4.6*10-6*0.2*10-6/0.05
= -165.23*10-3 + 413.08*10-3
= 0.247 J
(2) Initial potential energy= 8.98*109 *4.6*10-6*0.1*10-6/0.0514 + 8.98*109 *4.6*10-6*0.1*10-6/0.0514 +P.E due toq3q4
= 0.0803 + 0.0803 +P.E due toq3q4
= 0.16 J +P.E due toq3q4
Final potential energy = 8.98*109 *4.6*10-6*0.1*10-6/0.0233 + 8.98*109 *4.6*10-6*0.1*10-6/0.0233 +P.E due toq3q4
= 0.354 J +P.E due toq3q4
=> change in potential energy = 0.354 - 0.16
= 0.194 J
(3) Potential energy of system = 2*8.98*109 *4.6*10-6*0.1*10-6/0.0233 + 8.98*109 *0.1*10-6*0.1*10-6/0.024
= 0.354 + 0.00374
= 0.3577 J
(4) Potential energy of system = 0 - 8.98*109 *0.1*10-6*0.1*10-6/0.024
= - 0.00374 J
(5) Initial potential energy = 8.98*109 *4.6*10-6*0.2*10-6/0.05 + 8.98*109 *4.6*10-6*0.2*10-6/0.02 +P.E due toq2q6
= 0.165 + 0.413 +P.E due toq2q6
= 0.578 J +P.E due toq2q6
Final potential energy = 8.98*109 *4.6*10-6*0.2*10-6/0.02 + 8.98*109 *4.6*10-6*0.2*10-6/0.05 +P.E due toq2q6
= 0.413 +0.165 +P.E due toq2q6
= 0.578 J +P.E due toq2q6
=> Change in potential energy = 0 J
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