# Electric field and ball of charge

1. May 16, 2007

### Rasine

As shown in the figure above, a ball with a mass of 0.180 g and positive charge of q=31.0
mC is suspended on a string of negligible mass in a uniform electric field. We observe
that the ball hangs at an angle of q=16.0o from the vertical. What is the magnitude of
the electric field?

so i drew a free body diagram and now i am going to calcuate the forces in the x and y
directions to see what E is.

since the ball is stationary Fx=0=E-Tsin(16)+q

so E=Tsine(16)+q

now i solve for T by finding the forces acting in the y direction Fy=0=Tcos(16)-mg

so T=mg/cos(16)...so i substitute that into Fx and now E=(mg/cos(16))sin(16)+q

what am i doing wrong?

2. May 16, 2007

### Staff: Mentor

Good.

The force exerted by the (presumably horizontal) electric field on the charge is Eq. Rework this part.

3. May 16, 2007

### neutrino

What's with the '+q' term?

4. May 16, 2007

### Rasine

so it would be E=Tsin(16)+Eq?

5. May 16, 2007

### neutrino

No. It would be Tsin(16)-Eq = 0.

6. May 16, 2007

### Staff: Mentor

Not quite. Just redo your calculation of Fx.

7. May 16, 2007

### Rasine

Fx=0=E-Tsin(16)+the charge of the partical

where is the Electric field coming into play agian....i don't understand

8. May 16, 2007

### Staff: Mentor

This makes no sense. You are adding the horizontal forces on the charged object. What forces act on it? E is the field, not the force! q is the charge, also not a force.

Reread the posts by neutrino and myself.

9. May 16, 2007

### Rasine

so the force due to the electric field is F=Eq so instead of E=Tsine(16) i would have Eq=Tsin(16)??

10. May 16, 2007

### Staff: Mentor

Yes. Now you've got it.

11. May 16, 2007

### Rasine

thank you so much. i really appriciate it.