What is the magnitude of the electric field in a given scenario?

[physgrl]

Homework Statement

Refer to the figure below. Find the magnitude of the electric field if the string with a 7 gram ball carrying a charge of 4 x 10-9 C forms an angle of 35o with the vertical.


http://www.montereyinstitute.org/courses/AP%20Physics%20B%20II/course%20files/assignments/lesson31selfcheckquiz_files/image001.gif

a. 1.41 x 106 N/C
b. 1.20 x 107 N/C
c. 2.30 x 107 N/C
d. 6.25 x 108 N/C

Homework Equations

Fg=mg
E=F/q

The Attempt at a Solution

Force due to electric field=Force due to gravity:

F=mgcos(θ)
F=7*10^-3kg*9.81m/s^2*cos(35)
F=Eq
E=F/q
E=(7*10^-3kg*9.81m/s^2*cos(35))/(4x10^-9)C
E=1.41x10^7 N/C

which is not an answer...the correct answer in the key is b, but I don't know why?

 

[technician]

The force due to gravity is not equal to the force due to the electric field. It would help to draw the forces acting on the object.
There is a tension in the string and TCos35 is the vertical force = weight of ball
T Sin35 is the horizontal force = force due to electric field.
I got T = 0.084N which gave horizontal Force = 0.048N
Which gives (b) as the answer... see if you can check this out

 

[I like Serena]

Hi physgrl!

How did you get the formula F=mgcos(θ)?
I'm afraid it is not the horizontal force due to gravity.

 

[physgrl]

ohh, how can i get the tension force? I thought id't be the hypotenuse of downward weight but that would be .007*9.81/cos(35)=0.084

 

[I like Serena]

ohh, how can i get the tension force? I thought id't be the hypotenuse of downward weight but that would be .007*9.81/cos(35)=0.084

This is correct.
That is the tension force.

 

[technician]

That is correct... that is what I got
so the horizontal component of this tension force is TSin35 = 0.084Sin35 = 0.048N

 

[physgrl]

Ohh ok thanks! :)