A point charge q 2 = -2.6 ?C is fixed at the origin of a co-ordinate system as s

Question

A point charge q2 = -2.6 ?C is fixed at the origin of a co-ordinate system as shown. Another point charge q1 = 0.5 ?C is is initially located at point P, a distance d1 = 5.4 cm from the origin along the x-axis

Not sure how to answer 2-5. I know pythagorean theorem should be used but I'm a bit lost on how to apply it.

Potential Energy of Point Charges 23 4 5 6 A point charge q2 =-2.6 C is fixed at the origin of a co-ordinate system as shown. Another point charge q1 = 0.5 C is is initially located at point P, a distance d1 = 5.4 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 = 2.2 cm from the origin along the x-axis as shown? 0.3151515151 Submit Your submiss-0.3151515151 Computed value: -0.3151515151 Submitted: Sunday, January 17 at 9:01 PM Feedback: Your answer has been judged correct; the exact answer is: 0.315151515151515 2) a- q4 The charge q2 is now replaced by two charges q3 and q4 which each have a magnitude of-1.3 pC half of that of q2. The charges are located a distance a = 1.3 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? Submit 3) 13 d2 R 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. Submit

Explanation / Answer

1. For PE we use Ep =kq1*q2/r

=9 * 10^9 * q1 * q2 / r
PE = final - initial = 9*10^9 * 0.5* -2.6 * 10^-12 [ 1/0.022 - 1/0.054]
Because of negative sign, we have
PE = 9 * 0.5 * 2.6 * (1/54 - 1/22) joule =0.315 J Ans.

2. When finding the potential energy of the system we can add the separate contributions from each of the charge-charge interactions. This problem is just like the previous one except that, because of the y displacement, we have to use the Pythagorean Theorem to find the distances. Also in this particular case since Q3 and Q4 have the same charge and distances from Q1 we can just find the contribution from one of them and double it.

so revised distance d1 will be by pythogorus theorem = 5.5542 cm

and revised d2 = 2.5553 cm

hence PE = final - initial = 9*10^9 * 0.5* -2.6 * 10^-12 [ 1/0.025553 - 1/0.055542]
Because of negative sign, we have
PE = 9 * 0.5 * 2.6 * (1/55.542 - 1/25.553) joule =0.2473 J Ans.

3. Just use our potential energy equation from before and sum up the different contributions

U = kq1*q3/2.553 + kq1*q4/2.5553 + kq1*q2/1.3

U = 9*10^9 * 0.5* -2.6 * 10^-12 [ 1/0.025553] + 9*10^9 *-1.3* -1.3 * 10^-12 [ 1/0.0015]

= - 0.457 +10.14 =9.683 J .............Ans.

4. Because Q3 and Q5 have the same magnitude of charge although the charge is opposite and are at the same distance from Q1 the their interactions with Q1 cancel out and we only have to look at the potential between Q3 and Q5.

U = kq3*q5/1.3

U = 9*10^9 * 1.3* -1.3 * 10^-12 [ 1/0.0015]

= - 10.14 J .............Ans.

5. We go from a system where, with Q2 and Q6 having the same charge, we go from a system where one charge is at a distance of d2 and the other at d1, to a system where one charge is d1 distant and the other d2. Essentially nothing changes and the difference in potential is zero.

5.

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