The drawing shows two situations in which charges are placed on the x and y axes

Question

The drawing shows two situations in which charges are placed on the x and y axes. They are all located at the same distance of 8.10 cm from the origin O. For each of the situations in the drawing, determine the magnitude of the net electric field at the origin.

The drawing shows two situations in which charges are placed on the x and y axes. They are all located at the same distance of 8.10 cm from the origin O. For each of the situations in the drawing, determine the magnitude of the net electric field at the origin. E = E =

Explanation / Answer

The electric field magnitude = k*q/r^2

For a)

the 2 & -3?C charges create fields in the + x direction and the -5?C creates a field in the + y direction
so in the x ....
Ex = k*(q1/r1^2 + q2/r2^2) = 9.0x10^9*(2.0x10^-6/0.081^2 + 3.0x10^-6/0.081^2) = 6.85x10^7N/C
Ey = 9.0x10^9*5.0x10^-6/0.081^2 = 6.85x10^7N/C

So E = sqrt(Ex^2 + Ey^2) = sqrt((6.85x10^7N/C)^2 +(6.85x10^7N/C)^2) = 9.68x10^7N/C

b) In this case the 4 & -1 ?C charges create fields in the + x direction, the +1 creates a field in the - y and the 6 creates a field in the + y direction

So Ex = k/r^2*(q1 + q2) = 9.0x10^9/(0.081^2)*(4x10^-6 + 1.0x10^-6) = 6.85x10^7N/C

Ey = k/r^2*(6x10^6 - 1.0x10^6) = 9.0x10^9/(0.081^2)*(6x10^-6 - 1.0x10^-6) = 6.85x10^7N/C

So E = sqrt(Ex^2 + Ey^2) = sqrt((6.85x10^7N/C)^2 +(6.85x10^7N/C)^2) = 9.68x10^7N/C

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