Question

A parallel-plate capacitor is located horizontally so that one of its plates is submerged into a liquid while the other is over its surface. The permittivity of the liquid is equal to $ϵ$. Its density is equal to $\rho$. To what height will the level of the liquid in the capacitor rise after its plate get a charge of surface density $\sigma$ ?

• A. No change in liquid level
• B. $\frac{{\rho }^2({\in }^2-1)}{{\in }_0\rho g}$
• C. $\frac{{\rho }^2({\in }^2-1)}{2{\in }_0\rho g{\in }^2}$
• D. $\frac{{\rho }^2(\in -1)}{{\in }_0\rho g}$

Answer 1

The correct option is C $\frac{{\rho }^2({\in }^2-1)}{2{\in }_0\rho g{\in }^2}$


When the capacitor gets charged electric field will be induced between the plates. Since water is dielectric, charge will gets distributedd in it.
As we know the total charge induced in the dielectric is
${\sigma }^{\prime }=\sigma (1-\frac{1}{\in })$
Assuming the top plate is positively charged, there will be electric field from top to bottom. Simillarly electric field in water will be from bottom to top.
The electic field due to water will pull the water surface upward due to which the water level rises to some height $h$
Now calculating net electric field ,
$E_1=\frac{\sigma }{2{\in }_0}$ (downward)
$E_3=\frac{{\sigma }^{\prime }}{2{\in }_0}=\frac{\sigma (1-\frac{1}{\in })}{2{\in }_0}$ (upward)
$E_2=\frac{\sigma }{2{\in }_0}$ (downward)
$E_{Net}=(E_1+E_2)-E_3=\frac{\sigma }{2{\in }_0}+\frac{\sigma }{2{\in }_0}-=\frac{\sigma (1-\frac{1}{\in })}{2{\in }_0}$
$=\frac{\sigma }{{\in }_0}[1-\frac{1}{2}+\frac{1}{2\in }]=\frac{\sigma }{{\in }_0}[\frac{1}{2}+\frac{1}{2\in }]$
$E_{Net}=\frac{\sigma }{2{\in }_0}[1+\frac{1}{\in }]$ (downward)
For calculating force,
$F=qE$ As, $q={\sigma }^{\prime }A$
$\frac{F}{A}={\sigma }^{\prime }E$
$=\sigma (1-\frac{1}{\in })\frac{\sigma }{2\in }[1+\frac{1}{{\in }_0}]$
$\frac{F}{A}=\frac{{\sigma }^2}{2{\in }_0}(1-\frac{1}{{\in }^2})=P$ (upward)
As this in equilibrium then,
$\rho gh=\frac{{\sigma }^2}{2{\in }_0}(1-\frac{1}{{\in }^2})$
$h=\frac{{\sigma }^2}{2\rho g{\in }_0}(1-\frac{1}{{\in }^2})=\frac{{\sigma }^2}{2\rho g{\in }_0}\left(\frac{{\in }^2-1}{{\in }^2}\right)$
For detailed solution watch the next video.