Question

A particle of mass 1kg and carrying positive charge 0.01 C is sliding down an inclined plane of angle ${30}^0$with the horizontal. An electric field E is applied to stop the particle. If the coefficient of friction between the particle an the surface of the plane is$\frac{1}{2\sqrt{3}}$, E must be:

• A. 1260 V/m
• B. 245 V/m
• C. $\frac{1000\sqrt{3}}{7}V/m$
  
• D. $\frac{490}{\sqrt{3}}V/m$

Answer 1

Answer: (C)

$\frac{1000\sqrt{3}}{7}V/m$

In equilibrium

$N=Eqsin30+mgcos30-------\left(1\right)$ and

$f+Eqcos30=mgsin30-------\left(2\right)$

Maximum static function is $\mu N$

$\Rightarrow f=\mu N$

$\mu N+Eqcos30=mgsin30-------\left(3\right)$

Solving (1), (2) and (3)

$-\frac{Eqcos30}{\mu }+\frac{mgsin30}{\mu }=Eqsin30+mgcos30$

$E(\frac{qcos30}{\mu }+qsin30)=mg(-cos30+\frac{sin30}{\mu })$

$E\times (0\cdot 01)(\frac{\sqrt{3}\times 2\sqrt{3}}{2}+\frac{1}{2})=1\times 10(\frac{-\sqrt{3}}{2}+\frac{1}{2}2\sqrt{3})$

$E\times 0\cdot 01\times 5\frac{7}{2}=10\times \left(\frac{\sqrt{3}}{2}\right)$

$E=\frac{10\sqrt{3}}{0\cdot 01\times 7}=\frac{1000\sqrt{3}}{7}$