I am only looking for answers for Q3 to Q5. A point charge q 2 = -4.2 C is fixed

Question

I am only looking for answers for Q3 to Q5.

A point charge q2 = -4.2 C is fixed at the origin of a co-ordinate system as shown. Another point charge q1 = 4 C is is initially located at point P, a distance d1 = 7.7 cm from the origin along the x-axis

1)

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 = 3.1 cm from the origin along the x-axis as shown?

J

2)

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 oringin 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

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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

3.) The Potential energy of the system is nothing but the work done to get the system in that place.

So let us start arranging the charges one by one to get the PE of that system in Q3.

Since PE depends only on the final state, it doesn't matter in which order we place the charges.

Let me start with q1. To place q1 at point R, I need no energy since there is no potential as of now (because of the absence of any other charges in the environment in the beginning )

Next I will place q3 at 1.9 cm above the origin.

For that, I need to do some work which is = Potential at that point (due to charge q1) x charge placed

W3 = (1/4o) *(q1 / r13) x q3

r13 is the distance between q1 and q3 which is ( a2 + d22 ) 1/2 = ( 0.0312 + 0.0192 ) 1/2 = 0.036359 m

Next I need to place q4 at 1.9 cm below the origin

Work done to place charge 4 = Potential at point 4 (due to both the charges q1 and q3 ) x charge 4

W4 = ( (1/4o)*(q1/r14) + (1/4o) * (q3/r34) ) x q4  = (1/4o) x ( q1q4 / r14 + q3q4 / r34 )

So, net work done to bring the system to this state, i.e net PE of the system = W3 + W4

PE =  (1/4o) x ( q1q4 / r14 + q3q4 / r34 + q1q3 / r13 ) ............. eqn 0 (remember this formula for PE of system of 3 charges)

that's it, substitute q1 = 4 C, q3 = q4 = -2.1 C , r13 = r14 = 0.036359 m, r34 = 0.038 m

1/4o is approximately 9 x 109

Therefore, PE = 9x 109 x ( (4* - 2.1/ 0.036359 ) + (-2.1*-2.1/0.038) + (4 * -2.1 / 0.036359) ) x 10-12  (don't forget to convert all into SI units)

PE = - 3.114056528 Joules

Q4.) Use the equation we have derived for PE of system of three charges

PE =  (1/4o) x ( q1q4 / r14 + q3q4 / r34 + q1q3 / r13 ) ............. eqn 0

here replace q4 with q5

PE =  (1/4o) x ( q1q5 / r15 + q3q5 / r35 + q1q3 / r13 ) ............. eqn 0

don't change the other values except q5 = 2.1 C

PE = 9x 109 x ( (4* 2.1/ 0.036359 ) + (-2.1*2.1/0.038) + (4 * -2.1 / 0.036359) ) x 10-12

PE = - 1.044473684 Joules

Q5.) Let us use the same formula we have derived for PE of system of three charges

PE =  (1/4o) x ( q1q5 / r15 + q3q5 / r35 + q1q3 / r13 ) ............. eqn 0

but with charges q1, q2 and q6

PE =  (1/4o) x ( q1q2 / r12 + q2q6 / r26 + q1q6 / r16 ) ............. eqn 0

q1 = 4 C, q2 = q6 = -4.2C

Before moving the charge q1 , let the PE be Ui

before moving the charge r12 = d1 = 7.7 cm = 0.077 m, r26 = d1 + d2 = 10.8 cm = 0.108 m,

r16 = d2 = 3.1 cm = 0.031 m

Substituting these values in our formula, we get

Ui = 9x 109 x ( (4* -4.2/ 0.077 ) + (-4.2*-4.2/0.108) + (4 * -4.2 / 0.031) ) x 10-12    = - 5.371055718 J

After moving the charge q1 , let the PE be Uf

After moving the charge, r12 = d2 = 3.1 cm = 0.031 m, r26 = d1 + d2 = 10.8 cm = 0.108 m,

r16 = d - d2 = 10.8 - 3.1 cm = 0.077 m

Substituting these values in our formula, we get

Uf = 9x 109 x ( (4* -4.2/ 0.031 ) + (-4.2*-4.2/0.108) + (4 * -4.2 / 0.077) ) x 10-12    = - 5.371055718 J

PE = Uf - Ui = - 5.371055718 - (- 5.371055718) = 0 No change in PE

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