Question

A block of mass $m$ kept on a rough horizontal surface is connected to a dielectric slab of mass $m/6$ and dielectric constant $K$ by means of a light and inextensible string passing over a fixed pulley as shown in figure. The dielectric can completely fill the space between the parallel plate capacitor of plate width $l$. The separation between the plates is $d$ kept in vertical position. Initially switch $S$ is open and length of dielectric inside the capacitor is $b$. The coefficient of friction between the block and surface is $1/4$. The minimum value of emf of battery $V$, such that after closing the switch block start to move:
(Neglect any other forces)

• A. $\sqrt{\frac{mgd}{36{ϵ}_{0}lK}}$
• B. $\sqrt{\frac{2mgd}{3{ϵ}_{0}l(K-1)}}$
• C. $\sqrt{\frac{mgd}{6{ϵ}_{0}l(K-1)}}$
• D. $\sqrt{\frac{3mgd}{{ϵ}_{0}l(K-1)}}$

Answer 1

The correct option is C $\sqrt{\frac{mgd}{6{ϵ}_{0}l(K-1)}}$

The total forces acting on the dielectric are electrostatic attractive force due to the capacitor, tension due to the string and self weight.

The block will slip when tension exceeds the maximum friction value.

i.e., $T⩾(f_s{)}_{max}$


For equilibrium of slab,

$T=F_e+\frac{mg}{6}$

Thus, for minimum value of emf $V$, applying the just slipping condition on block.

$F_e+\frac{mg}{6}=(f_s{)}_{max}=\mu mg[\because T=\left(f_s{)}_{max}\right]$

$\Rightarrow F_e+\frac{mg}{6}=\frac{mg}{4}$

$\Rightarrow F_e=\frac{mg}{4}-\frac{mg}{6}=\frac{mg}{12}$

Force on the dielectric slab will be,

$F_e=\frac{l{ϵ}_0{V}^2(K-1)}{2d}(l=\text{plate width})$

$\therefore \frac{l{ϵ}_0{V}^2(K-1)}{2d}=\frac{mg}{12}$

$\Rightarrow$ $V_{min}=\sqrt{\frac{mgd}{6{ϵ}_{0}l(K-1)}}$

Hence, option $\left(c\right)$ is correct.

Why this question? Tip: In such problems you have to focus on F.B.D of body and include the effect of force on dieletric due to capacitor.