Energy of charged capacitor $$ = \frac{1}{2}C{V^2}$$
Energy given by cell $$ = C{V^2} = \frac{{A{\varepsilon _0}}}{d} \times {\left( {Ed} \right)^2}$$
As, $$V = Ed = A{\varepsilon _0}{E^2}d$$

$${C_{{\text{medium}}}} = K \times {C_{{\text{air}}}}$$

Let $${C_1} =$$  Capacity of capacitor with $${K_1}$$
$${C_2} =$$  Capacity of capacitor with $${K_2}$$
$${C_3} =$$  Capacity of capacitor with $${K_3}$$
$$\eqalign{ & \therefore {C_1} = {K_1}\left( {\frac{A}{2}} \right)\frac{{{\varepsilon _0} \times 2}}{d} = \frac{{A{\varepsilon _0}{K_1}}}{d} \cr & \therefore {C_2} = {K_2}\left( {\frac{A}{2}} \right)\frac{{{\varepsilon _0} \times 2}}{d} = \frac{{A{\varepsilon _0}{K_2}}}{d} \cr & \therefore {C_3} = {K_3}\left( A \right)\frac{{{\varepsilon _0} \times 2}}{d} = \frac{{2A{\varepsilon _0}{K_3}}}{d} \cr} $$
$${C_1}$$ and $${C_2}$$ are in parallel
$$\therefore {C_{eq}} = \frac{{A{\varepsilon _0}}}{d}\left( {{K_1} + {K_2}} \right)$$
$${C_{eq}}$$  and $${C_3}$$ are in series
$$\therefore \quad \frac{1}{C} = \frac{d}{{A{\varepsilon _0}\left( {{K_1} + {K_2}} \right)}} + \frac{d}{{2A{\varepsilon _0}{K_3}}}$$
But $$C = \frac{{KA{\varepsilon _0}}}{d}$$   for single equivalent capacitor
$$\eqalign{ & \therefore \frac{d}{{KA{\varepsilon _0}}} = \frac{d}{{A{\varepsilon _0}\left( {{K_1} + {K_2}} \right)}} + \frac{d}{{2A{\varepsilon _0}{K_3}}} \cr & {\text{or}}\,\frac{1}{K} = \frac{1}{{{K_1} + {K_2}}} + \frac{1}{{2{K_3}}} \cr} $$

Force between plates of parallel capacitor,
$$F = qE = q\left[ {\frac{\sigma }{{2{\varepsilon _0}}}} \right]$$
$$\because $$ Surface charge density $$\sigma = \frac{q}{A}$$
$$\therefore F = q\left[ {\frac{q}{{2A{\varepsilon _0}}}} \right] \Rightarrow F = \frac{{{q^2}}}{{2A{\varepsilon _0}}}$$
So, net charge across a capacitor, $$q = CV$$
$$F = \frac{{{C^2}{V^2}}}{{2A{\varepsilon _0}}}\,\,\,\,\,\left[ {C = \frac{{A{\varepsilon _0}}}{d}} \right]$$
$$ \Rightarrow F = \frac{{\left( {\frac{{A{\varepsilon _0}}}{d}} \right) \times C{V^2}}}{{2A{\varepsilon _0}}} = \frac{{C{V^2}}}{{2d}}$$

The figure shows two independent balanced wheatstone Brides connected in parallel each having a capacitance $$C.$$ So, $${C_{{\text{net}}}} = {C_{AB}} = 2\,C.$$

With the closing of switch $${S_3}$$ and $${S_1}$$ the negative charge on $${C_2}$$ will attract the positive charge on $${C_2}$$ thereby maintaining the negative charge on $${C_1}.$$ The negative charge on $${C_1}$$ will attract the positive charge on $${C_1}.$$ No transfer of charge will take place. Therefore p.d across $${C_1}$$ and $${C_2}$$ will be $$30 V$$  and $$20 V.$$

The common potential difference across two capacitors connected in parallel.
$${V_{{\text{eq}}}} = \frac{{{C_1}{V_1} + {C_2}{V_2}}}{{{C_1} + {C_2}}}$$
Here, potential of charged capacitor
$${V_1} = V,$$
potential of uncharged capacitor $${V_2} = 0$$
$$\therefore {V_{{\text{eq}}}} = \frac{{{C_1}V}}{{{C_1} + {C_2}}}$$

$$\eqalign{ & {\text{From}}\,{\text{figure,}}\,\frac{y}{x} = \frac{d}{a} \Rightarrow y = \frac{d}{a}x \cr & dy = \frac{d}{a}(dx) \Rightarrow \frac{1}{{dC}} = \frac{y}{{K{\varepsilon _0}adx}} + \frac{{\left( {d - y} \right)}}{{{\varepsilon _0}adx}} \cr & \frac{1}{{dC}} = \frac{y}{{{\varepsilon _0}adx}}\left( {\frac{y}{K} + d - y} \right) \cr & \int {dC} = \int {\frac{{{\varepsilon _0}adx}}{{\frac{y}{K} + d - y}}} \cr & {\text{or,}}\,C = {\varepsilon _0}a \cdot \frac{a}{d}\int\limits_0^d {\frac{{dy}}{{d + y\left( {\frac{1}{K} - 1} \right)}}} \,\,\,\,\,\,\left[ {\because dy = \frac{d}{a}dx} \right] \cr & = \frac{{{\varepsilon _0}{a^2}}}{{\left( {\frac{1}{K} - 1} \right)d}}\left[ {\ell n\left( {d + y\left( {\frac{1}{K} - 1} \right)} \right)} \right]_0^d \cr & = \frac{{K{ \in _0}{a^2}}}{{\left( {1 - K} \right)d}}\ell n\left( {\frac{{d + d\left( {\frac{1}{K} - 1} \right)}}{d}} \right) \cr & = \frac{{K{ \in _0}{a^2}}}{{\left( {1 - K} \right)d}}\ell n\left( {\frac{1}{K}} \right) = \frac{{K{ \in _0}{a^2}\ell nK}}{{\left( {K - 1} \right)d}} \cr & {\text{Alternatively remember}} \cr & C = \frac{{{K_1}{K_3}A{\varepsilon _0}}}{{d\left( {{K_1} - {K_2}} \right)}}{\log _e}\frac{{{K_1}}}{{{K_2}}} \cr} $$

$$\eqalign{ & {\text{Here}}\,{K_1} = 1,\,{K_2} = K,\,A = {a^2} \cr & \therefore C = \frac{{K{a^2}{\varepsilon _0}}}{{d\left( {1 - K} \right)}}{\log _e}\frac{1}{K} = \frac{{K{a^2}{\varepsilon _0}}}{{d\left( {K - 1} \right)}}{\log _e}K \cr} $$

Capacitance will increase but not $$5$$ times (because dielectric is not filled completely).
Hence, new capacitance may be $$200\,\mu F.$$

Let the level pf liquid at an instant of time $$'t'$$ be $$x.$$ Then
$$\eqalign{ & v = - \frac{{dx}}{{dt}} \Rightarrow dx = - vdt \cr & \Rightarrow \int\limits_{\frac{d}{3}}^x {dx = - v\int\limits_0^t {dt} } \cr & \Rightarrow x - \frac{d}{3} = - vt \cr & \Rightarrow x = \frac{d}{3} - vt \cr} $$

Also the capacitance can be considered as an equivalent of two capacitances in series such that
$$\eqalign{ & \frac{1}{{{C_{eq}}}} = \frac{1}{{{C_1}}} + \frac{1}{{{C_2}}} \cr & \Rightarrow \frac{1}{{{C_{eq}}}} = \frac{1}{{\frac{{{ \in _0}A}}{{d - x}}}} + \frac{1}{{\frac{{{ \in _O}AK}}{x}}} = \frac{{d - x}}{{{ \in _0}A}} + \frac{x}{{{ \in _o}AK}} \cr & \therefore {C_{eq}} = \frac{{{ \in _0}AK}}{{Kd + x\left( {1 - K} \right)}} \cr & {\text{But }}A = 1,K = 2{\text{ and }}x = \frac{d}{3} - vt \cr & \therefore {C_{eq}} = \frac{{{ \in _0} \times 1 \times 2}}{{2d + \left[ {\frac{d}{3} - vt} \right]\left( {1 - 2} \right)}} = \frac{{6{ \in _0}}}{{5d + 3vt}} \cr & \therefore {\text{Time constant }}\tau = R{C_{eq}} = \frac{{6R{ \in _0}}}{{5d + 3vt}} \cr} $$