In equilibrium state, the thread makes an angle of 60° with the vertical. The tension in the thread is resolved into horizontal and vertical components. Then, tension in the string in equilibrium, $\text{T}\cos60^\circ=\text{mg}$
$\text{T}\times\frac{1}{2}=\big(10\times10^{-3}\big)\times10$
$\text{T}=\big(10\times10^{-3}\big)\times10\times2=0.20\text{N}$
(b) As it is displaced from equilibrium, net force on the ball,$\text{F}=\sqrt{(\text{mg})^2+\Big(\frac{\text{q}\sigma}{2\in_0}\Big)^2}$
As F = ma$\Rightarrow\text{a}=\sqrt{(\text{g})^2+\Big(\frac{\text{q}\sigma}{m2\in_0}\Big)^2}$
The surface charge density of the plate (as calculated in the previous question), $\sigma=7.5\times10^{-7}\text{C/m}^2$ Charge on the ball, q = 4 × 10-6C Mass of the ball, m = The time period of oscillation of the given simple pendulum,$\text{T}=2\pi\sqrt{\frac{\text{l}}{\text{g}}}$
$=2\pi\sqrt{\frac{10\times10^{-2}}{9.8}}$
$=0.45\text{sec}$
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