Question

A small sphere of mass 5.10g is hanging vertically from an insulating thread that is 12.0 cm long. By charging some nearby plates, the sphere is subjected to a horizontal electric field of magnitude 7.20 N/C. As a result the sphere is displaced 6cm horizontally in the direction of the electric field.
(i) What is the angle that the thread makes with the vertical?
(ii) What is the tension in the thread?
(iii)What is the charge on the sphere?

Answer 1

The figure shows the condition of the problem and the symbols used. Triangle OAB is right with right angle OBA, the hypotenuse OA wich equals the length of thread - L=12.0 cm, and the legs AB wich equals the horizontal displacement d=6.0 cm of the sphere. Thus for the angle AOB (we denote $\alpha$) we can write $sin\alpha=d/L=6/12=1/2$ . One should know $\alpha=30\degree$ if $sin\alpha=1/2$. According to first Newton's law sum of all forces acts on the sphere must be 0. The force of weight acts on the sphere is $F_g=mg=5.10*10^{-3}[kg]*9.8 [m/s^2]=5.0*10^{-2}N$ . This force is balanced by $T_y$ (Look on the figure). $T_y$ is a leg of right triangle in wich the tension in the thread equals hypotenuse. Thus we have $T=T_y/cos\alpha=(note\space cos 30\degree=\frac {\sqrt {3}}{2})=\frac{2T_y}{\sqrt{3}}=5.77*10^{-2}N$

For the horizontal force we can write

$T_x=Tsin\alpha=\frac{T}{2}=2.89*10^{-2}N$ . The electric force must be equals to $T_x$ and acts on opposite direction. This is $F_e=E*q=T_x$ when $q$ is the charge on the sphere. Thus $q=\frac{T_x}{E}=\frac{2.89*10^{-2} N}{7.2 N/C}=4.0*10^{-3}C$ .

Answer:

(i) The angle that the thread makes with the vertical is $30\degree$

(ii) The tension in the thread is $5.77*10^{-2} N$

(iii) The charge on the sphere is $4.0*10^{-3}C$