Since the four charges are points charges, the equation for the field due to each individual charge is frequently written as
#E=(kQ)/r^2# where #k=9xx10^9(Nm^2)/C^2#
Now, we must calculate the field at P that is due to each charge separately, and finally add these contributions together (as vectors).
Since P is at the centre of a rectangle 20 cm long by 10 cm high, the distance #r# from any charge to P is found by Pythagorus' theorem:
#r^2 = (0.10)^2 + (0.05)^2 = 0.0125 m^2#
where you will note I have changed the distance into metres, to agree with the units of #k#.
Since #r^2# is the quantity we need, I shall not bother to find #r# itself!
Since #Q_1# and #Q_3# are equal, we can find the field due to these in a single calculation.
#E=(kQ)/r^2=((9xx10^9)(20xx10^(-6)))/0.0125 = 1.44xx10^7 N/C#
The #y#-components of these field vectors cancel (#E_(1y) = -E_(2y)#), and the #x#-component for both these fields is
#E_x=Ecos theta#
So, how to find #theta#? Basic geometry tells us that the angle of #E_1# is one that would have a tangent equal to #5-:10# (the same measures we used to find #r^2# above).
So, #theta=tan^(-1)(5/10) = 26.6°# and #costheta = 0.894#
Therefore, both #E_(1x)# and #E-(2x)# are given by
#E_x=1.44xx10^7 (0.894) = 1.29xx10^7 N/C# in the positive #x#-direction (away from the positive charge).
and the total field due to #Q_1# and #Q_3# is #2.58xx10^7 N/C#
As similar series of calculations to find #E_2# and #E_4# will yield the result the both #E_2# and #E_4# are equal in magnitude to #2.58xx10^7 N/C# (because the distances and angles are the same as for #E_1 and E_3# but the charges are twice as large.
So, the total #E_2+E_4# will equal #5.16xx10^7 N/C# and will also point in the positive #x#-direction (toward the negative charge).
Thus, the overall field is
#E_1+E_2+E_3+E_4 = 7.74xx10^7 N/C# in the positive #x#-direction.
