Question

A particle of mass 1 Kg carrying 0.01C charge is at rest on an inclined plane of angle ${30}^0$ with the horizontal when an electric field of $\frac{490}{\sqrt{3}}N{C}^{-1}$ is applied parallel to the horizontal as shown. The coefficient of friction is :

• A. $0.5$
• B. $\frac{1}{\sqrt{3}}$
• C. $\frac{\sqrt{3}}{2}$
• D. $\frac{\sqrt{3}}{7}$

Answer 1

The correct option is D $\frac{\sqrt{3}}{7}$

For the body to be at rest on the incline, all the forces acting on it should balance out each other.

Comparing forces along the incline , $qEcos30=\frac{490}{\sqrt{3}}\times \frac{\sqrt{3}}{2}\times 0.01=2.45N$

$mgsin30=1\times 9.8\times \frac{1}{2}=4.9N$

$\therefore qEcos30<mgsin30$

Hence , motion is downwards the plane and $f$ is up the inclined plane .
Force balance across the vertical direction gives

$mg=fsin\theta +Ncos\theta$ where $\theta ={30}^o$

Similarly, balancing forces along the x direction,

$Nsin\theta =fcos\theta +qE$

Now, if $\mu$ is the coefficient of friction, then $f=\mu N$
Substituting the above values in the equations and solving them simultaneously, we get the value of $\mu$ as $\frac{\sqrt{3}}{7}$