Question

Three positive point charges are located at the corners of a rectangle, as illustrated in the figure below. Find the electric field (in N/C) at the fourth corner if q1 = 4 nC, q2 = 6 nC, and q3 = 5 nC. The distances are d1 = 0.6 m and d2 = 0.2 m. (Let the positive x-direction be to the right, and the positive y-direction be upward.) Enet, x = N/C Enet, y N/C

A PHYS 1207 Homewerk e Clegg Study l Guided Solutions e Three Positive port Charges Are × 1.00 er iew-MATH 1036TL 12 Chttps.//www. t/Assignment Responses/submit?dep-20599863 8e-13 eBook 5. 114.28 points ZnFhysL53 18.P.018. My Notes Ask Your Three positive point charges are ocated at the corners of a rectangle, as llustrated in the figure below. Firnd the electrit field (irn N/C) at the ourth coner ifq-4 nC, d1-0.δ m and d2-0.2 m. (Let the positive x-dirction be to the right, and the positive y-direction be upward.) nC, and a- 5 nC. The distances are net, x Ni'C NIC rY eBook Subrit Aiswe Save Progress Practice Another Version 6. 14.2814.28 points | Prewous Answers ZinFhy9L53 16.P.023 My Notes Ask Your Al a cerlain loualion close lo Carh's surfa, we ubserve a ufor electric lield of magnilude 105 N/C dited slraighl down. Wha arder to make them lose cantact with the gMake sue to correctly identify the sign nf the charge needed. age (in C) hal needs lo be placed n a person of mass 92 ku in 01e-3 C. eBook

Answer 1

Electric field is given by:
E = k*q/r^2
Net electric field on 4th corner will be
Enet = E1 + E2 + E3
Since q1 , q2 and q3 are positive , So field will be repulsive.
Direction of E1 = force due to q1 = towards +y-axis in (north)
Direction of E2 = force due to q2 = towards -y-axis in (west)
Direction of E3 = Force due to q3 = at 45 deg below +x-axis (45 deg north of west)
Now
E1 = k*q1/r1^2
r1 = 0.2 m
Using given values
E1 = 9*10^9*4*10^-9/0.2^2
E1 = (900 N/C) j
Now
E2 = k*q2/r2^2
r2 = 0.6 m
Using given values
E2 = 9*10^9*6*10^-9/0.6^2
E2 = (150 N/C) -i

Now

E3 = k*q3/r3^2
r3 = sqrt(r1^2 + r2^2)

r3 = sqrt(0.2^2 + 0.6^2) = 0.63 m
Using given values
E3 = 9*10^9*5*10^-9/0.63^2
E3 = 113.38 N/C  

E3 = 113.38*(-cos(45 deg)) i + 113.38*(sin(45 deg)) j

E3 = -80.17 i + 80.17 j

So, Enet = E1 + E2 + E3

Enet = (-150 - 80.17)i + (900 +80.17) j

Enet = -230.17 i + 980.17 j

therefore,

Enetx = -230.17 N/C

Enety = 980.17 N/C

Please Upvote.