In: Physics
In the figure, four charges, given in multiples of 5.00Ý10-6 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 = 3.60Ý10-2 m. What are the magnitude and direction of the electric field at the center of the square? The magnitude of E? Ex? Ey?
a)
x direction:
Ex = (k q/d^2) (1 + 3/(2) * sqrt(2)/2 + 5/(2) * sqrt(2)/2 + 2 - 3/(2) * sqrt(2)/2 + 5/(2) * sqrt(2)/2)
Ex = (9e9*5e-6/(3.60e-2)^2) * (1 + 3/(2) * 0.7071 + 5/(2) * 0.7071 + 2 - 3/(2) * 0.7071 + 5/(2) * 0.7071)
Ex = 2.2693 x 10^8 N/C
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y direction:
Ey = (k q/d^2) (1 + 3/(2) * sqrt(2)/2 + 5/(2) * sqrt(2)/2 - 1 - 3/(2) * sqrt(2)/2 + 5/(2) * sqrt(2)/2)
Ey = (9e9*5e-6/(3.60e-2)^2) * (1 + 3/(2) * 0.7071 + 5/(2) * 0.7071 - 1 - 3/(2) * 0.7071 + 5/(2) * 0.7071)
Ey = 1.2276 x 10^8 N/C
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E = sqrt((2.2693x10^8)*(2.2693x10^8)+(1.2276x10^8)*(1.2276x10^8)) = 2.58 x 10^8 N/C
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b)
x direction:
Ex = (k q/d^2) (1 + 3/(2) * sqrt(2)/2 + 5/(2) * sqrt(2)/2 + 2 - 3/(2) * sqrt(2)/2 + 5/(2) * sqrt(2)/2)
Ex = (9e9*5e-6/(3.60e-2)^2) * (1 + 3/(2) * 0.7071 + 5/(2) * 0.7071 + 2 - 3/(2) * 0.7071 + 5/(2) * 0.7071)
Ex = 2.27 x 10^8 N/C
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c)
y direction:
Ey = (k q/d^2) (1 + 3/(2) * sqrt(2)/2 + 5/(2) * sqrt(2)/2 - 1 - 3/(2) * sqrt(2)/2 + 5/(2) * sqrt(2)/2)
Ey = (9e9*5e-6/(3.60e-2)^2) * (1 + 3/(2) * 0.7071 + 5/(2) * 0.7071 - 1 - 3/(2) * 0.7071 + 5/(2) * 0.7071)
Ey = 1.23 x 10^8 N/C