Question

pendulum of mass 80 mg and carrying a charge of 20nC is at rest in a horizontal uniform

electric field of 2 X10^4

Vm-1. Find the tension in the thread of pendulum and angle it makes

with the vertical.​

Answer 1

Answer:

0.8 N ; 0.286°

Explanation:

Gravitational force (vertical) $= mg = 80 \cdot 10^{-3} \cdot 10 = 0.8 \mathrm{N}$

Electrostatic force (horizontal) $= Eq = 2\cdot 10^4 \times 20 \cdot 10^{-9} = 4 \cdot 10^{-3} \mathrm{N}$

The resultant of these two forces is balanced by the tension in the string.

These forces are in perpendicular directions, so their resultant is given by:

$T = \sqrt{(mg)^2+(Eq)^2} = \sqrt{64\cdot10^{-2} + 16\cdot10^{-6}} \approx 0.8 \mathrm{N}$

Let angle made with vertical be $\theta$

$\tan \theta = \dfrac{Eq}{mg} = \dfrac{4 \cdot 10^{-3}}{8 \cdot 10^{-1}} = 5 \cdot 10^{-3}$

$\theta = \tan^{-1}(5 \cdot10^{-3}) \approx 0.286^{\circ}$