A point charge q2 = -3.7 ?C is fixed at the origin of a co-ordinate system as sh

Question

A point charge q2 = -3.7 ?C is fixed at the origin of a co-ordinate system as shown. Another point charge q1 = 1.8 ?C is is initially located at point P, a distance d1 = 8.6 cm from the origin along the x-axis

a) What is ?PE, the change in potenial energy of charge q1 when it is moved from point P to point R, located a distance d2 = 3.4 cm from the origin along the x-axis? J

b) The charge q2 is now replaced by two charges q3 and q4 which each have a magnitude of -1.85 ?C, half of that of q2. The charges are located a distance a = 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? J

c) What is the potential energy of the system composed of the three charges q1, q3, and q4, when q1 is at point R? Define the potential energy to be zero at infinity. J

d) The charge q4 is now replaced by charge q5 which has the same magnitude, but opposite sign from q4 (i.e., q5 = 1.85 ?C). What is the new value for the potential energy of the system? J

e) Charges q3 and q5 are now replaced by two charges, q2 and q6, having equal magnitude and sign (-3.7?C). Charge q2 is located at the origin and charge q6 is located a distance d = d1 + d2 = 12cm from the origin as shown. What is ?PE, the change in potential energy now if charge q1 is moved from point P to point R?

Explanation / Answer

A)

Potential enery at point P

PEP = kq1q2/d1

PEP = 9 x 109 x 1.8 x 10-6 x (-3.7 x 10-6) / 8.6 x 10-2

PEP = -0.7 J

Potential enery at point R

PER = kq1q2/d2

PER = 9 x 109 x 1.8 x 10-6 x (-3.7 x 10-6) / 3.4 x 10-2

PER = - 1.8 J

PE = PER - PEP

PE = -1.8 - (-0.7)

PE = -1.1 J

B)

Distance between P and q3 be

Rp = (d12 + a2)

Rp = (8.62 + 22)

Rp = 8.8 cm

Rp = 8.8 x 10-2 m

Distance between P and q4 would be equal to Rp

Initial PE of the system

PEp = k { (q1q3 /RP)+ (q1q4 /RP) + (q4q3 /2a) }

Since q3 = q4 = q = -1.83 x 10-6 C

PEp = k { (2q1q /RP) + (q4q3 /2a) }

Distance between R and q3 be

RR = (d22 + a2)

RR = (3.42 + 22)

RR = 3.9 cm

RR = 3.9 x 10-2 m

Distance between R and q4 would be equal to RR

Final PE of the system

PEp = k { (q1q3 /RR)+ (q1q4 /RR) + (q4q3 /2a) }

Since q3 = q4 = q = -1.83 x 10-6 C

PER = k { (2q1q /RR) + (q4q3 /2a) }

Chanrge in PE

PE = PER - PEP

PE = k { (2q1q /RR) + (q4q3 /2a) } - k { (2q1q /RP) + (q4q3 /2a) }

PE = 2kqq1{(1 /RR) - (1 /RP)}

PE = 2 x 9 x 109 x (-1.85 x 10-6) x 1.8 x 10-6 {(1/3.9 x 10-2) - (1/8.8 x 10-2)}

PE = - 0.86 J

C)

Final PE of the system

PER = k { (q1q3 /RR)+ (q1q4 /RR) + (q4q3 /2a) }

PER  = 9 x 109 { [ (1.8 x 10-6)x (-1.85 x 10-6)/3.9 x 10-2] +[(1.8 x 10-6)x(-1.85 x 10-6)/3.9x 10-2]+[(-1.85 x 10-6)x(-1.85 x 10-6)/2 x 2 x 10-2]}

PER = -0.77 J

D)

Final PE of the system

PER = k { (q1q3 /RR)+ (q1q5/RR) + (q5q3 /2a) }

PER  = 9 x 109 { [ (1.8 x 10-6)x (-1.85 x 10-6)/3.9 x 10-2] +[(1.8 x 10-6)x(1.85 x 10-6)/3.9x 10-2]+[(1.85 x 10-6)x(-1.85 x 10-6)/2 x 2 x 10-2]}

PER  = 0.9 { -0.854 + 0.854 - 0.855}

PER = - 0.77 J

E)

PEP = k { (q1q2 /d1)+ (q1q6 /d2) + (q2q6 /d) }
since q2 = q6 = q

PEP = k { (q1q /d1)+ (q1q /d2) + (q2 /d) }

PEP = kq { (q1/d1)+ (q1 /d2) + (q /d) }

PER = k { (q1q2 /d2)+ (q1q6 /d1) + (q2q6 /d) }

since q2 = q6 = q

PER = k { (q1q /d2)+ (q1q /d1) + (q2 /d) }

PER = kq { (q1/d1)+ (q1 /d2) + (q /d) }

Since PER = PEP

PE = 0

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