Question

Consider the situation of the previous problem. (a) Find the tension in the string in equilibrium. (b) Suppose the ball is slightly pushed aside and released. Find the time period of the small oscillations.

Answer 1

(a)
In equilibrium state, the thread makes an angle of 60^o with the vertical.
The tension in the thread is resolved into horizontal and vertical components.
Then, tension in the string in equilibrium,

$$\begin{gathered} T\mathit{}\cos 60°=m\mathrm{g} \\ T\times \frac{1}{2}=\left(10\times {10}^{-3}\right)\times 10 \\ T=\left(10\times {10}^{-3}\right)\times 10\times 2=0.20\mathrm{N} \end{gathered}$$

(b) As it is displaced from equilibrium, net force on the ball,

$F=\sqrt{{\left(mg\right)}^2+{\left(\frac{q\sigma }{2{\in }_0}\right)}^2}$
As F = ma
$\Rightarrow a=\sqrt{{\left(g\right)}^2+{\left(\frac{q\sigma }{m2{\in }_0}\right)}^2}$
The surface charge density of the plate (as calculated in the previous question), σ = 7.5×10^$-$7 C/m²
Charge on the ball, q = 4×10^$-$6 C
Mass of the ball, m =
The time period of oscillation of the given simple pendulum,

$$\begin{gathered} T=2\mathrm{\pi }\sqrt{\frac{l}{\mathrm{g}}} \\ =2\mathrm{\pi }\sqrt{\frac{10\times {10}^{-2}}{9.8}} \\ =0.45\sec \end{gathered}$$