Use Coulomb's law to determine the magnitude and direction of the electric field at points A and B in Fig. 16-

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