In the figure, four charges, given in multiples of 5.00Ý10-6 C form the corners of a...

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?

Solutions

Expert Solution

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

---------------------------

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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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

--------------------------------------------------------------------------------------------------------------------------------

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

genius_generous answered 1 week ago

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