Three spiders are resting on the vertices of a triangular web. The sides of the

Question

Three spiders are resting on the vertices of a triangular web. The sides of the triangular web have a length of a = 0.63 m, as depicted in the figure. Two of the spiders (S1and S3) have +7.4

Three spiders are resting on the vertices of a triangular web. The sides of the triangular web have a length of a = 0.63 m, as depicted in the figure. Two of the spiders (S1and S3) have +7.4 mu C charge, while the other (S2) has What are the magnitude and direction of the net force on the third spider (S3)? Magnitude: Direction: (Angle counterclockwise from +x axis) What are the magnitude and direction of the net force on the third spider (S3) when it is resting at the origin? Magnitude: Direction: (Angle counterclockwise from +x axis)

Explanation / Answer

charge on S1 is q1 = +7.4 uC = +7.4 x 10^-6 C is at (a,a/2)

charge on S3 is q2 = +7.4 uC = +7.4 x 10^-6 C is at ((3)^1/2 x a/2,0)

charge on S2 is q3 = -7.4 uC = -7.4 x 10^-6 C is at (-a,-a/2)

the electric field due to charge q1 at S3 is

E1 = k x (q1/r1^2)

where k = (1/4pi x e_o) = 9 x 10^9 Nm^2/C^2 and r1^2 = [((3)^1/2 x a/2 - a)^2 + (0 - a/2)^2] m^2

the electric field due to charge q3 at S3 is

E2 = k x (q3/r2^2)

where r2^2 = [((3)^1/2 x a/2 + a)^2 + (0 + a/2)^2] m^2

the net electric field at S3 is

E = (Ex^2 + Ey^2)^1/2

where Ex = E1x + E2x = E1 x cosA + E2 x cosB

and Ey = E1y + E2y = E1 x sinA + E2 x sinB

where the angles are

A = tan^-1(0 - a/2/(3)^1/2 x a/2 - a)

and B = tan^-1(0 + a/2/(3)^1/2 x a/2 + a)

the force on spider S3 is

F = E x q2

the direction of the net electric field is

theta = tan^-1(Ey/Ex)

the direction of the force is same as direction of net electric field

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