In the figure, four charges, given in multiples of 4.00×10 -6 C form the corners

Question

In the figure, four charges, given in multiples of 4.00×10-6 C form the corners of a square and four more charges lie at the midpoints of the sides of the square. The distance between adjacent charges on the perimeter of the square is d = 8.00×10-3 m.
What are the magnitude and direction of the electric field at the center of the square?

The magnitude of E?

What is the X component of the electric field (Ex) at the center of the square?

What is the Y component of the electric field (Ey) at the center of the square?

Explanation / Answer

to due to diagonal 3q charges, magnitude of field will be same but direction will be opposite

hence net field due to these two charges will be zero.

same is with +q charges.

magnitude and direction will be same for 5q and -5q.

now magnitude of E due to 5q :

E1 = k Q / r^2 = (9 x 10^9 x 5 x 4 x 10^-6 ) / (d^2 + d^2)

      = 1.406 x 10^9 N/C

due to 5q and -5q = E5 = 2 x 1.406 x 10^9 ) ( cos45i + sin45j)

      = (2i + 2j) x 10^9 N/C

due to q and -2q

E2 = (k q / d^2 ) + (k (2q) / d^2) = 3 k q / d^2

E2 = (3 x 10^9 x 9 x 4 x 10^-6 ) / (8 x 10^-3)^2

E2 = 1.688 x 10^9 N/C i

Enet = E5 + E2 = (3.7i + 2j) x 10^9 N/C

magnitude = sqrt(3.7^2 + 2^2) x 10^9 = 4.2 x 10^9 N/C

direction = tan^-1(2 / 3.7) = 28.4 deg

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Ex = 3.7 x 10^9 N/C

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Ey = 2 x 10^9 N/C

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