Secure https://session masteringp itemView? assignment Probleml D 753191008 offs

Question

Secure https://session masteringp itemView? assignment Probleml D 753191008 offset prev PHYS 212: Physics for Science & En ring II Spring '16 Problem 234 previous l 1 of 8 l next Problem 23.4 Part A What is the strength of the electric field at the position indicated by the dot in (Figure 1)? Submit My Answers Give up Part B of 1 What is the direction of the electric field at the position indicated by the dot in (Eigure 1) 3.0 nC Specify the direction as an angle measured counterclockwise from the positve z axis 5.0 cm 10 cm y Submi My Answers -6.0 nC Continue 21/2017

Explanation / Answer

formula for electric field due to point charge is

E = kq/ r^2

the electric field due to 3.0 nc charge is

E 1= 9 * 10^9 ( 3.0 * 10^-9)/((0.05)^2 = 10800 N/C i

from the figure

tan theta = 10 cm/ 5 cm

theta = tan^-1 ( 10/5) = 63.43 degree

electric field due to -6 nC charge is

E2 = -kq/ d^2 cos theta i - kq/d^2 sin theta j

here d^2 = ( 0.01)^2 + (0.05)^2

E2 = - 9* 10^9* ( 6* 10^-9)/( 0.01)^2 + (0.05)^2 * cos 63.43 - 9* 10^9* ( 6* 10^-9)/( 0.01)^2 + (0.05)^2 * sin 63.43

=- 9289.88 i - 18575.76 j

E total = E1 + E2

= 10800 N/C i-9289.88 i )+ (- 18575.76 j)

=1510.12 i -18575.76 j

magnitude

E = sqrt Ex^2 + Ey^2

= sqrt ( 1510.12)^2 + (-18576.76)^2

=1.863 * 10^4 N/C

direction

theta = tan^-1 ( -18576.76/1510.2) =- 85.35 degree

or 360-85.3 = 274.65 degree from the + x axis ( counter clock wise from + x axis )

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