Calculating the Angle of a Hanging Ball in an Electric Field

[cse63146]

Homework Statement

An electric field 3.00×10^5 N/C causes the ball in the figure to hang at an angle. What is the angle

Homework Equations

F_g = mg

The Attempt at a Solution

Here's a free body diagram that I drew in paint:

http://img339.imageshack.us/img339/6721/fbdnc7.jpg

I got f_g = mg = (0.002)(9.8) = 0.0196N = T_y

so would T_x be equal to E?

 

[rock.freak667]

The force exerted by the electric field on the 25nC ball is equal to T_x. Use F=EQ to find the electric force exerted on the ball.

 

[cse63146]

So:

F = EQ = (3x10^5)(25x10^-9) = 0.0075 N

then I just use tangent

tan$$\vartheta$$ = $$\frac{T_y}{T_x}$$ = $$\frac{0.0196}{0.0075 }$$ = 2.61

Does anyone see a mistake somewhere I made?

 

[hage567]

So:

F = EQ = (3x10^5)(25x10^-9) = 0.0075 N

then I just use tangent

tan$$\vartheta$$ = $$\frac{T_y}{T_x}$$ = $$\frac{0.0196}{0.0075 }$$ = 2.61

Does anyone see a mistake somewhere I made?

Check your trig. You've got Tx and Ty flipped.

Don't forget to actually calculate the angle.

 

[cse63146]

the 2.61 is after I used arctangent.

So let's try this one more time

tan$$\vartheta$$ = $$\frac{T_x}{T_y}$$ = $$\frac{0.0075}{0.0196}$$ = 20.9 (after arctangent was used).

This sounds more reasonable.