Question

What are the strength and direction of the electric field at the position indicated by the dot in (Figure 1)? 

Part A 

Give your answer in component form. (Assume that x-axis is horizontal and points to the right, and y-axis points upward.) 

Part B 

Give your answer as a magnitude and angle measured from the positive x-axis. 

Answer 1

Electric field is given by:

E = kq/r^2

direction of E will be towards -ve charge and away from +ve charge.

from FBD:

Angle A = arctan (2.0/4.0) = 26.57 deg

Now electric field E1 due to charge q1 will be towards the charge towards -ve y-axis

Electric field E3 due to charge q3 will be away from charge towards -ve x-axis

Electric field E2 due to charge q2 will be away from the charge, at 26.57 deg North of West

E1 = k*q1/r1^2

r1 = 2.0 cm = 0.02 m

E2 = k*q2/r2^2

r2 = sqrt (a^2 + b^2) = sqrt (0.02^2 + 0.04^2) = 0.0447 m

E3 = k*q3/r3^2

r3 = a = 4.0 cm = 0.04 m

Now Net electric field will be given by:

Enet = Ex_net i + Ey_net j

Ex_net = -E3 - E2x = -E3 - E2*cos 26.57 deg

Ex_net = k*q3/r3^2 - k*q2*cos 26.57 deg/r2^2

Ex_net = 9*10^9*10*10^-9/0.04^2 - 9*10^9*10*10^-9*cos 26.57 deg/0.0447^2

Ex_net = -96535.96 N/C

Ey_net = E2y - E1 = E2*sin 26.57 deg - E1

Ey_net = k*q2*sin 26.57 deg/r2^2 - k*q1/r1^2

Ey_net = 9*10^9*10*10^-9*sin 26.57 deg/0.0447^2 - 9*10^9*5.0*10^-9/0.02^2

Ey_net = -92352.67 N/C

So,

E_net = (-96535.96 i - 92352.67 j) N/C

In two significant figures

E_net = (-9.7*10^4 i - 9.2*10^4 j) N/C

Part B.

Magnitude of electric field is given by:

|E_net| = sqrt ((-96535.96)^2 + (-92352.67)^2)

|E_net| = 133597.18 N/C

|E_net| = 1.3*10^5 N/C

Now direction of electric field will be:

direction = arctan (-96535.96/(-92352.67)) = 46.27 deg below -ve x-axis

Direction = 180 + 46.27 = 226.27 = 230 deg counter-clockwise from +ve x-axis

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