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.
