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 from (Assume that x-axis is horizontal and points to the right and y-axis points upward.) Express your answer in terms of the unit vectors i and j. Express your answer using two significant figures. Give your answer as a magnitude and angle measured cw from the positive x-axis Express your answer using two significant figures. Express your answer using two significant figures.

Explanation / Answer

electric field at a point due to a point charge at distance r is E = kq/r^2 [ from coloumb's law]
so,
given, q1 = 5 nC
distance from the point P,r1 = 0.02 m
so vector field due to this charge at point P
E1 = kq1/r^2 i

similiarly
q2 = 10 nC
r2 = sqroot(0.02^2 + 0.04^2) = 0.0447 m
E2 = kq2(0.02i - 0.04j)/r2^3

similiarly
q3 = -5 nC
r3 = 0.04 m
E3 = -kq3 j/r3^2

Net field
E = E1 + E2 + E3 ( principle of superposition)
so,
E = k[q1/r^2 i + q2(0.02i - 0.04j)/r2^3 - q3 j/r3^2]
E = 8.98[5/0.02^2 i + 10(0.02i - 0.04j)/(0.0447)^3 + 5 j/(0.04)^2]
a. E = (17864.27)i - (40217.37)j V/m

b. in magnitude and angle notation
Magnitude, |E| = sqroot(Ex^2 + Ey^2) = 44006.46 V/m
direction ,. theta = arctan(Ey/Ex) = -66.04 deg CCW to +ve x axis

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