point charge q 2 = -4.2 C is fixed at the origin of a co-ordinate system as show
ID: 1355870 • Letter: P
Question
point charge q2 = -4.2 C is fixed at the origin of a co-ordinate system as shown. Another point charge q1 = 1.1 C is is initially located at point P, a distance d1 = 7.7 cm from the origin along the x-axis
The charge q2 is now replaced by two charges q3 and q4 which each have a magnitude of -2.1 C, half of that of q2. The charges are located a distance a = 1.9 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
3)
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
4)
The charge q4 is now replaced by charge q5 which has the same magnitude, but opposite sign from q4 (i.e., q5 = 2.1 C). What is the new value for the potential energy of the system?
J
5)
Charges q3 and q5 are now replaced by two charges, q2 and q6, having equal magnitude and sign (-4.2C). Charge q2 is located at the origin and charge q6 is located a distance d = d1 + d2 = 10.8cm 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?
J
Explanation / Answer
Q2 = -4.2 uC q1 = 1.1 uC r = 7.7cm
PE = k x Q1 x Q2/r ( k = 9 x 109 N-m-m/C,constant)
= 9 x 109 x(-4.2 x 10-6) x 1.1x10-6 / (7.7 x 10-2) = 0.54 J
When Q1 is replaced with q3(-2.1) and q4(-2.1) we will have 3 pairs q2q3, q4q3 and q2q4 and will calculate all PEs
When q2 is moved to R the distance d2 = 10.8-7.7 = 3.1
distance between q2q3 and q2q4 = 3.63 cm
Total PEq3q4 = 9x109 ((-2.1 x -2.1x 10-12)/1.9 x 10-2 = 2.088 J
PE q2q3 =9 x 109 x (-2.1) x 1.1x 10-12/3.63 x 10-2 = -0.572 J
PE q2q4 = 9 x 109 x (-2.1) x 1.1x 10-12/3.63 x 10-2 = -0.572 J
Total PE =0.105 J
Chnage in PE = 0.54 - 0.1 = 0.44 J
When q4 is replaced with q5 (2.1 uc)
PE q2q5 = 0.572 J
PE q3q5 = 2.088
Total PE = 2.088 +0.572 -0.572 = 2.088 J
when q4 and q5 are replaced with q2(-4.2) and q6(-4.2)
PE q2q6 = 9x 109 x(-4.2 x-4.2)x10-12 /10.8 x 10-2 = 0.147 J
PE q1q2 = 9 x 109 x(-4.2x1.1)x10-12/7.7 x 10-2 = -0.54 J
PE q1q6 = 9 x 109 x(-4.2x1.1)x10-12/3.1 x 10-2 = -0.134 J
Total PE = 0.147 -0.54 -0.134 = -0.527 J
When q1 is moved to R (3.1 cm) PE of pairs q1q2 and q1q6 will interchange
There will be no change in Total PE of the system
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