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1. Electric Charges and Fields
normal
A charged cork of mass $m$ suspended by a light string is placed in uniform electric filed of strength $E= $$(\hat i + \hat j)$ $\times$ $10^5$ $NC^{-1}$ as shown in the fig. If in equilibrium position tension in the string is $\frac{{2mg}}{{(1 + \sqrt 3 )}}$ then angle $‘\alpha ’ $ with the vertical is
A
$60^o$
B
$30^o$
C
$45^o$
D
$A$ and $B$ both
Solution
$E_{x}=E_{y}=10^{5} \mathrm{NC}^{-1}$
$q E_{x}=T \sin a$
$q E_{y}=m g-T \cos a$
Dividing $\frac{E_{x}}{E_{y}}=\frac{T \sin a}{m g-T \cos a}$
or $m g-T \cos a=T \sin a$
or $m g=\frac{2 m g}{1+\sqrt{3}}(\cos a+\sin a)$
or $\operatorname{ocsa}+\sin a=\frac{\sqrt{3}+1}{2}, \Rightarrow a=30^{\circ}$ or $60^{\circ}$
Standard 12
Physics
