91影视

Given $胃=17.4掳.m=12.3gandQ=-1.11渭C$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

$T\cos \left(17.4掳\right)=mg$鈥︹�︹�� (1)

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

$T\sin \left(17.4掳\right)=F_a=EQ$鈥︹�︹�︹��. (2)

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

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

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

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.