Solve step 4

.' Chrome File Edit View History Bookmarks Profiles Tab Window Help Thu Feb 16 5:15 PM . O M HW4_Sprin92023_Electric Fis X n Course Hero X i + v 6 C' i webassign.net/web/Student/Assignment-Responses/submit?dep=31402671&tags:autosave#question4272803_4 3 ] 1': I I] Update 3 Three charges are at the corners of an equilateral triangle, as shown in the gure below. Calculate the electric field at a point midway between the two charges on the x-axis. (Let q1 = 5.75 pC, q2 = 8.25 pc, and 43 = 8.25 uC.) y '7. 0.500 m 60.0 1: ti. 0. (a Slep1 The electric fields El, 32, and 23 at point P resulting from the charges ql, '72! and :13, respectively, are shown in the diagram. Since q2 is a positive charge, its electric field, 22, is directed away from P along the x-axis. The electric field E3, resulting from the negative charge q3, is also directed away from P along the xaxis toward q3. Point P is the midpoint of the base of the equilateral triangle. A line drawn from ql, at the top of the triangle, to P is perpendicular to the x-axis, so the electric field E1 at P resulting from positive charge q1 is perpendicular to the x-axis, directed away from P in the negative y-direction. The altitude h of the triangle and the equal distances from P to q2 and P to q3 are shown in the diagram. (D Since the altitude h of the triangle is perpendicular to its base and the triangle is equilateral, n = (0.500 \\J/ 7/ 0.5 m)sin(| y ,1 60 ) = y ,J 0.433 m. Step 2 Chrome File Edit View History Bookmarks Profiles Tab Window Help Q 8. Thu Feb 16 5:15 PM VA HW4_Spring2023_Electric Fiel X Course Hero X + C webassign.net/web/Student/Assignment-Responses/submit?dep=31402671&tags=autosave#question4272803_4 Update : Step 2 The magnitude of E, is given by 191 IE1 1 = 61 = Ke 7 2' where ke is Coulomb's constant and r, is the distance, h, of P from q, . We have 1911 = Ke - E1 = Ker 2 7 2 5.75 9 5.75 x 10-6 C) = (8.99 x 109 N . m2/C2) 0.433 0.433 m = 2.76 2.76 x 105 N /C . The vector E, is along the negative y-direction so it has no horizontal component, that is, Ex = 0, and, for its vertical component, Ely, we have Ely = -E1 = - 2.76 2.76 x 105 N/C. Step 3 The vectors E, and E3 are both along the positive x-direction so each has no y-component, that is, Ezy = 0 and Ey = 0. In the following equations, 2 is the distance of 92 from P and 3 is the distance of 93 from P. Thus, for the x-components of E2 and E3, Ezx and E3x , respectively, we have 192 E2x = Ker 2 = (8.99 x 109 N . m2 /C3)( x 10-6 c) (0.250 m) 2 x 105 N /C , and 1931 E3x = Ker.2 = (8.99 x 109 N . m2 /C2) ( x 10-6 c) (0.250 m)2 x 105 N/C. Submit Skip (you cannot come back).Chrome File Edit View History Bookmarks Profiles Tab Window Help 0 7 9 8. Thu Feb 16 6:34 PM VA HW4_Spring2023_Electric Fiel X Course Hero X + C webassign.net/web/Student/Assignment-Responses/submit?dep=31402671&tags=autosave#question4272803_4 Update The vectors E2 and E3 are both along the positive x-direction so each has no y-component, that is, Ezy = 0 and Ezy = 0. In the following equations, 2 is the distance of 92 from P and 3 is the distance of 93 from P. Thus, for the x-components of E2 and E3, Ezx and E3x , respectively, we have 192 Ezx = Ker,2 = (8.99 x 109 N . m2 /C2) (8.25 8.25 x 10-6 c) (0.250 m) 2 = 11.86 11.9 x 105 N /C , and 1931 E3x = Ker,2 = (8.99 x 109 N . m2 /C2) (8.25 8.25 x 10-6 C (0.250 m) 2 = 11.86 11.9 x 105 N/C. Step 4 The vector sum of the three vectors E,, E2, and E3, resulting in vector E, is represented in the diagram below. We obtain the x-component Ex of the resultant vector E by summing the x-components of E1 , E2, and E3- Ex = E1x + Exx + E3x = 0 + E2x + E3x x 106 N/C 0.500 m 60.0 O-- 121