Question

A point charge q1 = -3.00 nC is at the point x = 0.600 m, y = 0.800 m, and a second point charge q2 = 6.00 nC is at the point x = 0.600 m, y = 0. Calculate themagnitude and direction (counterclockwise from the x-axis is positive) of the net electric field at the origin due to these two point charges.

Answer 1

Distance of q1 from origin=r1= sq rt(0.6)^2 + (0.8)^2=1 m

Distance of q2 from origin=r2=0.6 m

Electric field due to q1 =E1=kq1/r1^2

Electric field due to q1 =E1=9*10^9*3*10^-9/1^2

Electric field due to q1 =E1=27 N/C at angleO with x axis,where tanO=53 degree

Component of E1 along x axis =27cos53=16.24 N/C (i)

Component of E1 along y axis =27sin53=21.56N/C (j)_______________________________
Electric field due to q2 =E2=kq2/r2^2

Electric field due to q2 =E2=9*10^9*6*10^-9/0.6^2

Electric field due to q2 =E2=150 N/C along (-) x axis

In vector notation,E2 =150N/C(-i)
_____________________________

Net field = E = E1+ E2

E = 16.24 N/C (i) + 21.56 N/C (j) + 150N/C(-i)

E = 133.76 N/C (-i) + 21.56 N/C(j)

Magnitude of resultant = E =sq rt 18356.57 =135.49 N/C

Angle O which Emakes with negative x axis is given by tanO= 21.56/133.76=0.16118

Angle O =9.156 degree clockwise with negative x axis or 189.156 degree clockwise with positive x axis

Answer 2

you need to use geometry to find the direction of the vector and the distance, and then use the following:

Electric field
E = F/Q = kQ/r²
in Newtons/coulomb OR volts/meter
k = 1/4πε₀ = 8.99e9 Nm²/C²

But your question is confusing. your first question states you want the field due to one charge, but then you list two charges. But if there are two charges, you just calculate the field due to each and add them with vector arithmetic.