An inclined plane making an angle $30°$with the horizontal is placed in a uniform horizontal electric field E of $100 V{m}^{-1}$ as shown. A particle of mass 1 kg and charge 0.01 C is allowed to slide down from rest from a height of 1 m, If the coefficient of friction is 0.2, find the time, in millisecond it will take the particle to reach the bottom. $\text{ (Take }\mathrm{g}=10{\mathrm{ms}}^{-2}\text{ ) }$

The different forces on the particle are shown in figure

From figure,

$\mathrm{N}=\mathrm{mgcos}{30}^{\circ }+\mathrm{qEcos}{60}^{\circ }$

By definition, function is

$f=\mu N=\mu mg\cos {30}^2+\mu qE\cos {60}^0$

If a is the acceleration of the particle down the incline, then

$mg\sin {30}^0-\mu N-qE\cos {30}^0=ma$

$\Rightarrow ma=mg\sin {30}^0-mg\cos {30}^0-\mu qE\cos {60}^0-qE\cos {30}^0$

Thus acceleration of the particle down the inciline is

$a=g\sin {30}^0-\mu g\cos {30}^0-\frac{\mu qE}{m}\cos {60}^0-\frac{qE}{m}\cos {30}^0$

$\Rightarrow a=10\left(\frac{1}{2}\right)-\left(0.2\right)\left(10\right)\left(\frac{\sqrt{3}}{2}\right)-\frac{\left(0.2\right)\left(0.01\right)\left(100\right)}{1}\left(\frac{1}{2}\right)-\frac{\left(0.01\right)\left(100\right)}{1}\left(\frac{\sqrt{3}}{2}\right)$

$\Rightarrow a=5-\sqrt{3}-0.1-\frac{\sqrt{3}}{2}$

$\Rightarrow a=2.3m{s}^{-2}$

Now, distance travelled in time t is

$s=0+\frac{1}{2}a{t}^2$

$\Rightarrow t=\sqrt{\frac{2\times 2}{a}}$

$\Rightarrow t=\sqrt{\frac{4}{2.3}}$

$\Rightarrow t=1.32s$

$\Rightarrow t=1320\times {10}^{-3}s$

$\Rightarrow t=1320ms$