Question

The bob has a mass 0.8 kg and a charge 1  C. If
an electric field (2ˆi  4ˆj) ×106 N/C is produced in
the region, angle made by the string with vertical
in equilibrium will

Answer 1

Given that,

Mass of bob = 0.8 kg

Charge =1 μC

Electric field $E=(2i+4j)\times10^{-6}\ N/C$

As the bob is in equilibrium

Net horizontal forces =0

$T\cos\theta=mg+qE_{y}$

$T=\dfrac{mg+qE_{y}}{\cos\theta}$...(I)

Net vertical forces =0

$T\sin\theta=E_{x}q$

$T=\dfrac{E_{x}q}{\sin\theta}$.....(II)

We need to calculate the angle made by the string with vertical

Using balance equation (I) and (II)

$\dfrac{mg+qE_{y}}{\cos\theta}=\dfrac{E_{x}q}{\sin\theta}$

$\tan\theta=\dfrac{E_{x}q}{mg+E_{y}q}$

Put the value in the equation

$\tan\theta=\dfrac{2\times10^{6}\times1\times10^{-6}}{0.8\times9.8+1\times10^{-6}\times4\times10^{6}}$

$\theta=\tan^{-1}(\dfrac{25}{148})$

$\theta=9.6^{\circ}$

Hence, The angle made by the string with vertical is 9.6°.