We have three forces acting on the ball: mg, string tension, and electric force.

The electric force must be repelling the ball away from the wall. We'll represent these forces in a free body diagram then decompose any 2D forces into their x and y components. Finally, We'll apply Newton's second law in the x and y-directions.

Newton's second law:

$\overline{){\mathbf{\Sigma}}{\mathbf{F}}{\mathbf{=}}{\mathbf{m}}{\mathbf{a}}}$

Electric force:

$\overline{){\mathbf{F}}{\mathbf{=}}{\mathbf{q}}{\mathbf{E}}}$

2D vector components:

$\overline{)\begin{array}{rcl}{\mathbf{T}}_{\mathbf{x}}& {\mathbf{=}}& \mathbf{\left|}\stackrel{\mathbf{\rightharpoonup}}{\mathbf{T}}\mathbf{\right|}\mathbf{}\mathbf{cos}\mathbf{}\mathbf{\theta}\\ {\mathbf{T}}_{\mathbf{y}}& {\mathbf{=}}& \mathbf{\left|}\stackrel{\mathbf{\rightharpoonup}}{\mathbf{T}}\mathbf{\right|}\mathbf{}\mathbf{sin}\mathbf{}\mathbf{\theta}\end{array}}$ where θ is the angle measured from the x-axis.

A small 12.8 g plastic ball is tied to a very light 25.2 cm string that is attached to the vertical wall of a room. (See the figure.) A uniform horizontal electric field exists in this room. When the ball has been given an excess charge of -1.20 μC, you observe that it remains suspended, with the string making an angle of 17.4° with the wall.

a) Find the magnitude of the electric field in the room.

b) Find the direction of the electric field in the room.

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