What are the strength and direction of the electric field at the position indica

Question

What are the strength and direction of the electric field at the position indicated by the dot in the figure(Figure 1)? Give your answer in component form. Assume that x- axis is horizontal and points to the right, and y-axis points upward. Express your answers using two significant figures separated by a comma. What is the magnitude of the electric field at the position indicated by the dot in the figure. Express your answer using two significant figures. What is the angle measured of from the positive x-axis of the electric field at the position indicated by the dot in the figure. Express your answer using two significant figures

Explanation / Answer

Using Coulomb's Law we find the magnitude of each field

So k*q/r^2 For +10nC field is in the - y direction so Ey = - 9.0x10^9*10x10^-9/0.03^2 = -1.00x10^5N/C

For the -5nC the field is in the +x direction and
Ex = 9.0x10^9*5x10^-9/0.05^2 = 18000N/C

for the last charge the distance is sqrt(0.03^2 + 0.05^2) = 0.0583
and the magnitude = 9.0x10^9^10x10^-9/0.0583^2 = 26470N/C but its components are
Ex = E*(5/5.83) = 26470*5/5.83 = 22700N/C
and Ey = E*3/5.83 = 26470*3/5.83 = 13620N/C

Now adding these we get Ex = 18000+22700 = 40700 and Ey = -1.00x10^5+13620 = -86400

So answer E = (4.1x10^4 i - 8.6x10^4 j )N/C

b) E = sqrt((4.1x10^4)^2 + (8.6x10^4)^2) = 9.5x10^4N/C

c) =arctan(-8.6x10^4/4.1x10^4) = -64.5o

so answer = -65o

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