For part(b), I don’t understand the answer. From the expression of VL and VQ got

Question

For part(b), I don’t understand the answer. From the expression of VL and VQ got from (a), in (b)(highlighted in yellow) why is rho the proportion of rho0 and rhoA? I think it should just be the distance between A(5,0,1) and the line (5,1,1). Also don’t understand why (1/r0 - 1/rA). thanks. 4.7 Electric Potential 143 Apoint charge of 5 nC is located at (-3,4,0), while line y-1. z = 1 carries uniform charge 2 nC/m. (a) If v =0vat 0(0,0,0), find vat A(5,0,1). (b) If V = 100Vat B( 1, 2,1), find Vata-2,5,3). (c) If V =_5 Vato,find v,c. Solution: XAMPLE 4.11A Let the potential at any point be where VQ and v, are the contributions to V at that point due to the point charge and the line charge, respectively. For the point charge, /E-d1 = -I4m,a,.th --9 For the infinite line charge, v. =-j E.dl=-jmpa,.dpa, 2T8. Hence, Q+C ATe 2TE where C= C' + C,-constant, is the perpendicular distance from the line y = 1. z 1 to the field point, and r is the distance from the point charge to the field point. (a) Ifv=oat 0(0, 0, o), and Vat A(5,0,1)isto be determined, we must first determine the values of andrat O and A. Finding r is easy; we use eq. (2.31) TO finde for any point (xy,s),we utilize thefact that is the perpendicular distance from (x, y, z) to line z=i, which is parallel to the x-axis. Hence is the distance between (x,y,z) and (x, 1,1) because the distance vector between the two points is perpendicular to a, Thus y . Applying this frand eq. (2.31)forat points O and A, we obtain Po - 1(o,0,0) - (0,1,1) - V2 ro-|(0,0,0)-(-3,4,0)| = 5

Explanation / Answer

In the given answer formula of line charge and point charge is used for finding potential between two points.

You are right, rho0 and rhoA is distance between origin and A for line charge and r0 and rA is distance between point charge.

V0=0 And VA we have to find, therefore rho0=(0,0,0)-(0,1,1) as y=1and z=1 is given for line charge, and rhoA=(5,0,1)-(5,1,1) .

Similarly for point charge r0=(0,0,0)-(-3,4,0) and rA=(5,0,1)-(-3,4,0).

And after putting all values in formula we get final answer.

Please forget whatever is given before part a solution in above picture, because for this kind of question you need only formula of particular charge, and after that you can find distance and final answer by putting in formula.

If you find any difficulty please find me in comment box.

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