Use Coulomb's law to determine the magnitude and direction of the electric field at points A and B in Fig. 16-57 due to the two positive charges (Q = 7.0 µC) shown. Is your result consistent with Fig. 16-31b?
magnitude at point A= ___ N/C
direction= ___° above the horizon
magnitude at point B= ___ N/C
direction= ___° above the horizon
Figure 16-57: http://i28.tinypic.com/2130i6e.gif
Figure 16-31b: http://i31.tinypic.com/2h4w3e9.gif
Please explain your answer and look up my other open physics questions. If possible, please help on those as well. Thanks!!
Use Coulomb's law to determine the magnitude and direction of the electric field at points A and B in Fig. 16-
Point A
In the x direction
r1=sqrt(25+100)/100 m
r2=sqrt(25+100)/100 m
at Point A the fields in the x cancel.
Let's look at y
C*Q*2*(100^2/125)/sqrt(2) upward
711.8*10^4 N/C in the positive y (90 degrees above the horizon)
Point B
in the x direction
r1=5*sqrt(2)/100 m
r2=sqrt(25+15^2)/100 m
from the left
C*Q*(1/r1^2)/sqrt(2)
from the right
-C*Q*(1/r2^2)*0.15/r2
add them together
Ex=C*Q*((1/r1^2)/sqrt(2)-
(1/r2^2)*0.15/r2)
6509847 N/C in the plus x direction
now the y
from the left
C*Q*(1/r1^2)/sqrt(2)
from the right
C*Q*(1/r2^2)*0.05/r2
add them together
Ey=C*Q*((1/r1^2)/sqrt(2)+
(1/r2^2)*0.05/r2)
9693014.1 upward
The resultant is
sqrt(Ex^2+Ey^2)
11676156.5 N/C
direction is atan(Ey/Ex)
56.11 degrees above the horizontal
________________________-
Reply:Use trig to figure the distances and angles of A %26amp; B from the 2 charges Q.
For each point (A %26amp; B) find the net field along the direction from the charge. Add these fields as vectors at each point by adding the x %26amp; y components of each field. The result will be your answer.
En = k*q/d² (k = 9E9)
There's a lot of busy work here, which why I'm giving general directions only.
ada
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