Question

Three charges are positioned at three corners of a square of side length a = 0.5 m as shown in the figure. The charges have values +2Q, -Q, and +Q, as shown, where Q = 7 μC. What is the y-component of the force on the charge +2Q? a. F = -1.14 N b. F = 4.77 N c. F = -1.76 N d. F = -4.77 N e. F = -2.28 N What is the work required to create this charge configuration? Assume the charges are brought in from infinity, where the electric potential is zero, and that these are the only charges in the universe. a. W = -0.881 J b. W = -2.79 J c. W = 3.89 J d. W = -1.4 J e. W = -3.89 J What is the electric potential at point P (the fourth vertex of the square)? a. V = 2.02 V b. V = -3.99 × 105 V c. V = 2.89 × 105 V If you bring an additional negative charge -2Q from infinity to point P, the work you do is a. zero. b. positive. c. negative

Three charges are positioned at three corners of a square of side length a = 0.5 m as shown in the figure. The charges have values +29,-Q, and +Q, as shown, where Q = 7 °C. +29 Р a - + a 1) What is the y-component of the force on the charge +2Q? a. F=-1.14 N b. F = 4.77 N c. F=-1.76 N d. F= -4.77 N e. F= -2.28 N 2) What is the work required to create this charge configuration? Assume the charges are brought in from infinity, where the electric potential is zero, and that these are the only charges in the universe. a. W = -0.881 J b. W = -2.79 J c. W = 3.89 d. W=-1.4 J e. W = -3.89 J 3) What is the electric potential at point P (the fourth vertex of the square)? a. V = 2.02 V b. V=-3.99 x 105 V c. V = 2.89 x 105 v 4) If you bring an additional negative charge -2Q from infinity to point P, the work you do is a. Zero. b. positive. c. negative.

Answer 1

Lesious Fone on +22 by -Q, F = kx2QxQ (towards -Q) az 45 B a Fusys YF. a k=9x109 Q => me = 7810 7x106c SATA a=0.5m. F = kx2Q ² fore Sch and 2Q by tQ is resolved caso KX2Q ² cos componean. Ag²=alta²=20² 2=k+QQ xQ - KQ² ABZ zaz Ey = kx Kx Q² XL a2 V2 92 (0.52 Ner y-component of total sone, y = Esin 45-5, K+20² - KQ KQ" [ta - -] - 9xio%26x700) (711293) Fy = -2.2810 [option e Work required to create the configuranon Pfennial energy (eleemly of sysko (PE tk(Q)C-Q) + Kx 2QxQ LAB²=20² W=PE = KA2Q XC-Q) ra AB=ha. Kad (-2+14) = 9:00'a(20-4) -1:58] 0.5 W=-1.45 [option d] v=kx2Q +K(-Q) a 0.5 Potennal ar P. Jza 9x10x7x10-6 2, 293] =2.89x109 [option c] 4) Work done to bring to P = its poreanal energy qu W = -2Q+ 2.89 4105=-ve value 19=-20 Coption c] negasive bring -20