A small 12.3-g plastic ball is tied to a very light 28.6-cm string that is attached to the vertical wall of a room (Fig. 1). A uniform horizontal electric field exists in this room. When the ball has been given an excess charge of -1.11 mC, you observe that it remains suspended, with the string making an angle of 17.4° with the wall. Find the magnitude and direction of the electric field in the room.

Fig.1

Given $\theta =17.4°.m=12.3gandQ=-1.11\mu C$and

We have to find Electric field and its direction.

As shown in the figure the ball is in equilibrium so the vertical component of tension is in balance with the weight so

$T\cos \left(17.4°\right)=mg$……… (1)

And the force due to the electric field is balanced with the horizontal component of tension

$T\sin \left(17.4°\right)=F_a=EQ$…………. (2)

From (1) and (2), we get that

localid="1665131389377" $$\begin{gathered} E=\frac{mg\sin \left(17.4°\right)}{Q\cos \left(17.4°\right)}=\frac{12.3\times {10}^{-3}\times 9.8\times \tan \left(17.4\right)}{1.11\times {10}^{-6}} \\ E=3.41\times {10}^{4}N/C \end{gathered}$$

The electric field is $3.41\times {10}^{4}N/C$

Hence the charge on the ball is negative and the force is to the right which means we can say there is a positive charge somewhere in the room so the field lines are out of positive and are oriented to negative so the field is to left.