Question

3. A-6uC charge and a 3C charge are placed 1.5 meters apart, with the - 6uC charge on the left side and the 3C charge on the right side. Use figure with arrow lines, as well as mathematical calculation to determine the following: a. (5 pts) Will you be able to place a -2uc charge outside of the 6C charge so that the net force from coming from the 3C and -6uC charges add up to be zero? If not, why? If so, where is the spot? b. (5 pts) Will you be able to place a -2uc charge outside of the 3C charge so that the net force from coming from the 3uC and -6uC charges add up to be zero? If not, why? If so, where is the spot? C. (5 pts) Will you be able to place a -2uc charge between the - 6uC and 3u charges so that the net force from these two charges add up to be zero? If not, why? If so, where is the spot?

Answer 1

31c 3) -CMC ? cosm (a) het the -rue change in pleu at a distance from bec. - 2 ис -Chic 300 ofr 1.5m F sé force due to buc = kq, 92 h :Ě K(-6X166) 1-2x166) F = (-) 2 X © に (-i) force due to 3u C k (3x16%)(-2x106) (x+1.5)2 Fi + F 0 = K (12x102) (1) - K (6 x16"} (x +15) Fret = Fit z? 2 11 2 (x+1:5)? Х x x + 1.5 xlL 2 X-X -1.5 da x 1:52 12-1

a we Sinia can not be negative have taken ill an positive quantity Hence this condition is not posible. (b) - Inc 3116 21 > Fi 1.5m ܪܐܢ; force due to - GMC ~> Free Fi = K (-6x106)(-2x106) in (oe+115) force due to suc F = K (3x10*')(-2x106) x2 K (12x166) fi - K(6x10') i (x+15)2 : Fnet K(12x166) K (6x10-6 (ve+1.5) (x+1.5) 2x ? x + 1.5 = x x2 x2 1.5 - 0.414 x = 1.5 0.414 -3.6 m X x m Since x is prexitive value Hence this condition is possible. (c) -luc -2uc F, 34c Afis 1.5m force due to - 60C k (-6x106)(-2x1024) i x2 Fi K(12 X10'22 i x2 force due to 3 MC F K ( 3x1056) (-2x106) 69) - (15-4)2

F = K (6x1042) (1.5 -12 Fret = K (12x1072) it ki6x16'?) (1.5-x)? clearly we can see fret can never be zero. Hence this Condition is not possible. *२