Question

In free space, a point charge Q1=9eQ1=9e CC is located at (0,0,0)(0,0,0) and a point charge Q2=−eQ2=−e CC is located at (10,0,0)(10,0,0).a) Find the potential at point PP (3,−2,8)(3,−2,8) due to these two point charges.V=V=  VVb) It is known that the equipotential surface having V=0V=0 VV is a spherical surface. Find the location of the origin of the equipotential surface having V=0V=0 VV.O=O= ((  ,  ,  ))c) Find the radius RR of the equipotential surface having V=0V=0 VV.R=R=

At least one of the answers above is NOT correct. 1 point) In free space, a point charge Q1 = 9e C is located at (0, 0, 0) and a point charge Q2--c C is located at (10,00 a) Find the potential at point P (3,-2,8) due to these two point charges. V1.3426 10 V b) It is known that the equipotential surface having V 0Vis a spherical surface. Find the location of the origin of the equipotential surface having V-0V 125.0 c) Find the radius R of the equipotential surface having V 1.25 0V

Answer 1

Solution: a) Given Q: ne at (0,0,01 Qise at (10,0,0) Here e is change on electron e: 16X169c Potential at point P(3,-2,8) Distance between P md Q, à d. 13-0376-2-03 + (8-oj? = 3+2781 - 877 Distance between P and Q, ' d2 = √3-10)+(-2-0) + (8-0) - V7+2 787 - |117 = 10.82 Potential p due to Q, and Q₂ in Voice on the release [9x16x1019 16X10 En 8.77 10.82 = q*v8x16x17" Le boi] [vision and 13.44 x 100v.

(b) K(C): Equipotential surface having U=0 Set (2,9,2) le the point where = 0 = qe , 4760 Vaxty' √2-10) tytz Potential due to Q. Potential due to Q2 at (1,4,7) at (x, y, z) hana Cherepot benista] Chess multipleying and squaring gives 81{(x-10% + y + z ] = x+y+z² 80x'7804't 807- 16204 + 8100 = 0 -) 2?+47 2- 20-25x + 101-25=0 (x-10.125)*7 y'+z" = 102:51-101-25 - 1.266 The above equation representa candle with Centre at (10.125,0,0) and radius 11.266 = 1.125 .: 10= (10230,0]], R = 1125]