Question

Consider the situation shown in figure (6−E9). Suppose a small electric field E exists in the space in the vertically charge Q on its top surface. The friction coefficient between the two blocks is μ but the floor is smooth. What maximum horizontal force F can be applied without disturbing the equilibrium?
[Hint The force on a charge Q bye the electric field E is F = QE in the direction of E.]

Answer 1

Explanation:

It is given that the force F = QE and we can write, $\mathrm{R}_{1}+\mathrm{QE}-\mathrm{mg}=0$,

Where, Q is vertical charge and E is Electric field.

So, $R_{1}=m g-Q E$, this can be considered as First equation.

We know that $F-T-\mu R_{1}=0$, where μ is the friction coefficient.

So, we can write $F-T-\mu m g+\mu Q E=0$, this can be considered as Second equation.

$\mathrm{R}_{1}+\mathrm{QE}-\mathrm{mg}=0$

$R_{1}=m g-Q E$ ....(1)

$\mathrm{F}-\mathrm{T}-\mu \mathrm{R}_{1}=0$

$\Rightarrow \mathrm{F}-\mathrm{T} \mu(\mathrm{mg}-\mathrm{QE})=0$

$\Rightarrow \mathrm{F}-\mathrm{T}-\mu \mathrm{mg}+\mu \mathrm{QE}=0$ ....(2)

$\mathrm{T}-\mu \mathrm{R}_{1}=0$

$\Rightarrow T=\mu R_{1}=\mu(m g-Q E)=\mu m g-\mu Q E$

From first and second equation, we can write $F=2 \mu(m g-Q E)$

Now the equation 2 is $\mathrm{F}-\mathrm{mg}+\mu \mathrm{QE}-\mu \mathrm{mg}+\mu \mathrm{QE}=0$

$\Rightarrow \mathrm{F}-2 \mu \mathrm{mg}+2 \mu \mathrm{QE}=0$

So, the maximum horizontal force that can be applied without disturbing the equilibrium is $2 \mu(m g-Q E)$.