# Electric Field and Force

## 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

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 dont know why?

Related Introductory Physics Homework Help News on Phys.org
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
Homework Helper
Hi physgrl! How did you get the formula F=mgcos(θ)?
I'm afraid it is not the horizontal force due to gravity.

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
Homework Helper
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.

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

Ohh ok thanks!! :)