A square 1 m on its side has charges on every corner. Calculate the electric field on the upper right corner of the square. Let the charge on each corner be 40 x10-9C.

Do I use K(Q*4)/r^2=F
Then F/(Q*4)=E
K=8.99*10^9 Nm^2/C^2

No, no, no.

Forces are vectors, you have to add them as vectors.

You have the across the diagonal force (distance is not 1), and the two corner forces at 90 degrees to each other. If you analyse it you can eliminate all side directions, so (In my head) the force will be along the diagonal direction
F=kq^2(1/s^2)(sin45 + cos45+1/2)

where s is 1 m

To calculate the electric field on the upper right corner of the square, you can use Coulomb's Law. The formula for electric field (E) is given by E = F / Q, where F is the force exerted on the charge and Q is the charge itself.

The formula you mentioned, F = K(Q * 4) / r^2, is the formula for the electric force (F) between two charges. Here, K represents the electrostatic constant (8.99 x 10^9 Nm^2/C^2), Q represents the charge on one corner (40 x 10^-9 C), and r represents the distance between the charges.

However, since we are calculating the electric field at a specific point due to the charges on the other corners, we need to consider the contributions of all the charges. To do this, we can calculate the electric field due to each individual corner charge and add them up.

In this case, the electric field due to each corner charge is given by E = KQ / r^2. Since the distance (r) between the charge and the upper right corner is the same for all four charges, we can simply calculate the electric field due to one charge and then multiply it by four.

Using the formula E = KQ / r^2, with K = 8.99 x 10^9 Nm^2/C^2, Q = 40 x 10^-9 C, and r being the side length of the square (which is 1 m), we can calculate:

E = (8.99 x 10^9 Nm^2/C^2 * 40 x 10^-9 C) / (1 m)^2

Simplifying this equation gives us:

E = 8.99 x 40 N/C

E = 359.6 N/C

So, the electric field at the upper right corner of the square is 359.6 N/C.