Question

Two metallic plates 1 m wide are kept 0.01 m apart like a parallel plate capacitors in air in such a way that one of their edges is perpendicular to the oil surface in a tank filled with an insulating oil. The plates are connected to a battery of e.m.f 500 V. The plates are then lowered vartically into the oil at a speed of $0.001m{s}^{-1}$. The current drawn from battery when dielectric constant of oil taken as 11 is :

• A. $4.43\times {10}^{-13}$
• B. $4.43\times {10}^{-8}$
• C. $4.44\times {10}^{-7}$
• D. $4.43\times {10}^{-9}$

Answer 1

The correct option is D $4.43\times {10}^{-9}$

THe adjacent figure in s given case,

$C_1+C_2$

$=\frac{K{\epsilon }_o(x+1)}{d}+\frac{{\epsilon }_o[(1-x)\times 1]}{d}$

where [K=Dielectric constant of oil]

$C=\frac{{\epsilon }_o}{d}[kx+1-x]$

After time $dt$ the dielectric rises by $dx$ the new equivalent capacitance will be

$C+dc=C_1^{\prime }+C_2^{\prime }$

$\frac{K{\epsilon }_o}{d}[(x+dx)\times 1]+\frac{{\epsilon }_o[(1-x-dx)\times 1]}{d}$

$=$Change of capacitance in time $dt$

$=\frac{{\epsilon }_o}{d}[kx+kdx+1-x-dx-kx-1-x]$

$\frac{{\epsilon }_o}{d}(k-1)dx$

$\frac{dc}{dt}=\frac{{\epsilon }_o}{d}(k-1)\frac{dx}{dt}=\frac{{\epsilon }_o}{d}(k-1)V$ $[$where $V=\frac{dx}{dt}]⟶(i)$

We know that,

$q=cv$

$\frac{dq}{dt}=V\frac{dc}{dt}⟶\left(ii\right)$

$I=V\frac{{\epsilon }_o}{d}(k-1)V$

From $\left(i\right)$ and $\left(ii\right)$

$I=\frac{500\times 8.85\times {10}^{-12}}{0.01}(11-1)\times 0.001$

$=4.43\times {10}^{-9}amp$