Please assist with non graded problem. Question about the item b of Chapter 4, Problem 21P – Sadiku's Elements of Electromagnetics 7th Edition (I have submited Bartleby's solution link but I was asked to sent another question without it. I did it, and was asked AGAIN to resubmit copying the question, so here it is): 21P: A ring placed along y2 + z2 = 4, x = 0 carries a uniform charge of 5 mC/m. a) Find D at P(3, 0, 0). b) If two identical point charges Q are placed at (0, -3, 0) and (0, 3, 0) in addition to the ring, find the value of Q such that D = 0 at P. ABOUT THE SOLUTION PROVIDED BY BARTLEBY: Shouldn’t r be calculated between each charge and P(3,0,0) – in that case, Bartleby is wrong? Please refer to the image for what I've done. What am I doing wrong? Can you further explain the progress made in the solution? Thank you!
Please assist with non graded problem.
Question about the item b of Chapter 4, Problem 21P – Sadiku's Elements of
21P: A ring placed along y2 + z2 = 4, x = 0 carries a uniform charge of 5 mC/m.
a) Find D at P(3, 0, 0).
b) If two identical point charges Q are placed at (0, -3, 0) and (0, 3, 0) in addition to the ring, find the value of Q such that D = 0 at P.
ABOUT THE SOLUTION PROVIDED BY BARTLEBY:
Shouldn’t r be calculated between each charge and P(3,0,0) – in that case, Bartleby is wrong? Please refer to the image for what I've done. What am I doing wrong? Can you further explain the progress made in the solution?
Thank you!
![Dtotal = D+ DQ, +DQ2
(128)
Charges field:
EQ
aR
(129)
4TE, R?
Q(r – rQ)
4TE,R3
(130)
in which ar is measured from the charge to P.
Therefore, the flux density for the charges is:
Q(r – rQ)
4|r – rq|3
(131)
For the charge put in (0, –3, 0), the flux density is:
Q (3,0,0) – (0, –3,0)
4т [(3,0, 0) — (0, —3, 0)|3
Q (3, 3,0)
4T (V18)3
(132)
(133)
For the charge put in (0,3,0), the flux density is:
Q (3,0,0) – (0, 3,0)
4т |(3,0, 0) — (0, 3, 0)[3
Q (3, –3,0)
4т (V18)3
Therefore, the flux in P shall be 0 and is given by:
Q (3, 3, 0)
(134)
(135)
Q (3, –3,0)
0 = 0.32a +
(136)
4т (V18)3
4x (V18)3
Q (6,0, 0)
4r (V18)³
6Q (1,0,0)
4# (V18)³
= 0.32(1,0, 0) +
(137)
= 0.32(1,0,0) +
(138)
Therefore:
6Q
-0, 32
(139)
4т(V18)3
which gives:
Q = -51, 182µC
(140)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F39066a3b-f737-4638-aede-b213a4f40f8c%2F0e7690df-2b9c-4b86-a8a9-dd020e222beb%2Fpdxwhlx_processed.png&w=3840&q=75)
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