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

The different forces on the particle are shown in figure:

From figure,
$\mathrm{N}=\mathrm{mg} \cos {30}^0+\mathrm{qE} \cos {60}^0$
By definition, friction is

${\text{⨍ =μN=μmg cos 30}}^0{\text{+μqE cos 60}}^0$

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

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

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

Thus, acceleration of the particle down the incline is

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

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

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

$\Rightarrow \mathrm{a}=2.3 {\mathrm{ms}}^2$

Now, distance travelled in time t is

$\mathrm{S}=0+\frac{1}{2}{\mathrm{at}}^2$

$\Rightarrow \mathrm{t}=\sqrt{\frac{2\times 2}{\mathrm{a}}} \left\{\because \mathrm{S}=\frac{1}{\sin 30}=2\right\}$

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

$\Rightarrow \mathrm{t}=1.32 \mathrm{s}$