If point charges of value +2[nC] are located at the points P 1 (2,2,-1)[m]and P

Question

If point charges of value +2[nC] are located at the pointsP1(2,2,-1)[m]and P2(1,-2,1)[m],respectively, find the voltage Vab betweenpoints Pa(2,3,1)[m] andPb(-1,2,2)[m] by using superposition principleand the potential expression for point charges V=Q/(4oR) [V] If point charges of value +2[nC] are located at the pointsP1(2,2,-1)[m]and P2(1,-2,1)[m],respectively, find the voltage Vab betweenpoints Pa(2,3,1)[m] andPb(-1,2,2)[m] by using superposition principleand the potential expression for point charges V=Q/(4oR) [V]

Explanation / Answer

You have all the variables that you need, although the problem isworded a bit oddly it is quite simple. Superposition simplystates that you can add the effects of different chargescharges. Vab is the difference in voltage of point a andpoint b. Q=+2[nC] Distance between points P1 and Pa = d1a Distance between points P1 and Pb=d1b Distance between point P2 and Pa=d2a Distance between point P2 and Pb=d2b V1a is the voltage from 1 on a. V1a=Q/(4**o*d1a) V2a is the voltage from 2 on a. V2a=Q/(4**o*d2a) Va=V1a+V2a V1b is the voltage from 1 on b. V1b=Q/(4**o*d1b) V2b is the voltage from 2 on b. V2b=Q/(4**o*d2b) Vb=V1b+V2b Finally; Vab=Va-Vb

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