In the figure below, q 1 = 1.70××1077 C and q 2 = 3.70××1077 C. In the figure be

Question

In the figure below, q1 = 1.70××1077 C and q2 = 3.70××1077 C.

In the figure below, q1 = 1.70 x 10 7 C and q2 = 3.70× 10 7 c. x (cm) 1 2 3 4 5 6 78 1) What is the x component of the electric field Eat the point (x, y) (6.00 m, 3.00 m)? (Express your answer to three significant figures.) NOTE the figure is correctly labeled in centimeters and the question is asking for the E-field at (6,3) meters. Account for this in your caículation. 89.4 Submit | You currently have 3 submissions for this question. Only 10 submission are allowed You can make 7 more submissions for this question Your submissions: | 89.4 . X Computed value: 89.4 Feedback Submitted: Monday, January 29 at 5:13 PM 2) what is the y component of the electric field E, at the point (x, y) = (6.00 m, 3.00 m)? (Express your answer to three significant figures.) 44.8 Submit You currently have 1 submissions for this question. Only 10 submission are allowed You can make 9 more submissions for this question. Your submissions:44.8X Computed value: 44.8 Feedback Submitted: Monday, January 29 at 5:13 PM 3) What is the x component of the force F acting on an proton at that position? (Express your answer to three significant figures.) x101 N Submit You currently have 0 submissions for this question. Only 10 submission are allowed You can make 10 more submissions for this question. 4) What is the y component of the force F acting on an proton at that position (Express your answer to three significant figures.) x1018 N Submit You currently have 0 submissions for this question. Only 10 submission are allowed You can make 10 more submissions for this question

Explanation / Answer

Given

charges are q1 = 1.7*10^-7 C at 0.05 cm on x-axis and

q2 = 3.7*10^-7 C at 0.08 cm on x-axis

and the point where the electric field to be determined at point P is (x2,y2) = (6m,3m)

we know that the electric field due to a charge q at r is , E = kq/r^2

here the distance is r = sqrt((x2-x1)^2+(y2-y1)^2)

electric fiedl due to q1 at P is E1 and electric fiedl due to q2 at P is E2

now electric field is a vector so writing in terms of components

E1 = E1x i + E1y j  

E1 = E1 cos theta i + E1 sin theta j

E1x = kq1/r1^2 * cos theta , cos theta = (6-0.05)/(sqrt(5.95^2+3^2)) ==> theta = 26.75 degrees

E1y = kq1/r1^2 * sin theta , sin theta = (3)/(sqrt(5.95^2+3^2)) ==> theta = 26.75 degrees

E1 = (9*10^9*1.7*10^-7 )/((sqrt(5.95^2+3^2)))^2 N/C = 34.458 N /C

now E1 = E1x i + E1y j  

E1 = E1 cos theta i + E1 sin theta j

E1 = 34.458*cos 26.75 i + 34.458*sin26.75 j

E1 = 30.77 N/C i + 15.51 N/C j

same way for field E2

E2 = E2x i + E2y j  

E2 = E2 cos theta i + E2 sin theta j

E2x = kq2/r2^2 * cos theta , cos theta = (6-0.08)/(sqrt(5.92^2+3^2)) ==> theta = 26.87 degrees

E1y = kq1/r1^2 * sin theta , sin theta = (3)/(sqrt(5.92^2+3^2)) ==> theta = 26.87 degrees

E2 = kq2/r^2 = (9*10^9*3.7*10^-7 )/((sqrt(5.92^2+3^2)))^2 N/C = 75.60 N /C

E2 = E2x i + E2y j  

E2 = 75.60 cos 26.87i + 75.60 sin 26.87 j

E2 = 67.42 N/C i + 34.17 N/C j

nwo the x component of the electric field is  

Ex = E1x+E2x = (30.77+67.42)N/C = 98.19 N /C and

2)

the y component of the electric field is  

Ey = E1y+E2y = (15.51+34.17)N/C = 49.68 N /C and

3) the force on proton is F = kq1*q/r1^2

F1 = F1x i + F1y j

F1 = (9*10^9*1.7*10^-7*1.6*10^-19 )/((sqrt(5.95^2+3^2)))^2 N = 5.51*10^-18 N

F2 = (9*10^9*3.7*10^-7*1.6*10^-19 )/((sqrt(5.92^2+3^2)))^2 N = 12.09*10^-18 N

F1 = F1 cos theta i + F1 sin theta j

F1 = 5.51*10^-18*cos26.75 i +5.51*10^-18*sin26.75 j = 4.92*10^-18 N i+2.48*10^-18 N j

and F2 = 12.09*10^-18 cos(26.87)N i + 12.09*10^-18 sin(26.87)N j= 10.78*10^-18 N i+5.464*10^-18 N j

now the x component of the force is  

Fx =(4.92*10^-18)+ 10.78*10^-18 N = 15.7*10^-18 N

4)

Fy =(2.48*10^-18)+ 5.464*10^-18 N = 7.944*10^-18 N

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