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 fiqure(Fiqure 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 answer in terms of the unit vectors i cap and j cap. 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

Sol.:- from the formula E = kq/d^2

E = 9x10^9 x Q/(d^2)

a) 5 nC particle

E1= 9x10^9 x Q/(d^2)

E1= 9x10^9 x (5x10^-9)/(0.02^2) = 112500 N/C (direction east theta1=0)

b) -5 nC particle

E2= 9x10^9 x Q/(d^2)

E2= 9x10^9 x (5x10^-9)/(0.04^2) = 28125 N/C (direction north theta2=90)

c) 10 nC particle

E3= 9x10^9 x Q/(d^2) ;;;;;; d = redical (2^2+ 4^2) = 4.472 cm (Pythagoras)

E3= 9x10^9 x (10x10^-9)/(0.044472^2) = 45000 N/C (direction: theta = -63.43)
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PART A:

Horizontal component : E x = 112500 + 45000.cos 63.43 = 132628.08 N/C
Vertical component : E y = 28125 - 45000 = -12122.48 N/C

====> E = 132628.08 i - 12122.48 j

PART B
Magnitude: E= redical (132628.08^2 +12122.48^2) = 133180.93 N/C
Angle: theta = tan-1 ( -12122.48/132628) = -5.22 degrees = 354.78 degree from + x (CW)

PART C
354.78 degree from + x axis (CW)

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