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 = 9 nC, q2 = 8 nC, and q3 = 7 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.)

42 d2 43 41

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*9*10^-9/0.2^2
E1 = (2025 N/C) j
Now
E2 = k*q2/r2^2
r2 = 0.6 m
Using given values
E2 = 9*10^9*8*10^-9/0.6^2
E2 = (200 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*7*10^-9/0.63^2
E3 = 158.73 N/C  

E3 = 158.73*(-cos(45 deg)) i + 158.73*(sin(45 deg)) j

E3 = -112.24 i + 112.24 j

So, Enet = E1 + E2 + E3

Enet = (-200 - 112.24)i + (2025 + 112.24) j

Enet = -312.24 i + 2137.24 j

therefore,

Enetx = -312.24 N/C

Enety = 2137.24 N/C

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