CBSE Questions for Class 12 Medical Physics Electrostatic Potential And Capacitance Quiz 9 - MCQExams.com

Given,

Both capacitors are in series

Capacitance of A & B, $${{C}_{A}}=2\mu F\,\,\And \,\,{{C}_{B}}=3\mu F$$

Total Potential is $${{V}_{T}}=({{V}_{A}}+{{V}_{B}})=10V$$

Let,

Potential difference on capacitor A & B, $${{V}_{A}},\,{{V}_{B}}$$  

In series, Capacitor have equal charge

$${{Q}_{A}}={{Q}_{B}}$$

$${{C}_{A}}{{V}_{A}}={{C}_{B}}{{V}_{B}}$$

$$\dfrac{{{V}_{A}}}{{{V}_{B}}}=\dfrac{{{C}_{B}}}{{{C}_{A}}}$$

$$\dfrac{{{V}_{A}}+{{V}_{B}}}{{{V}_{B}}}=\dfrac{{{C}_{B}}+{{C}_{A}}}{{{C}_{A}}}$$

$${{V}_{B}}=\dfrac{{{C}_{A}}({{V}_{A}}+{{V}_{B}})}{{{C}_{B}}+{{C}_{A}}}=\dfrac{2\times 10}{5}=4V$$

$${{V}_{B}}=4V$$

Similarly, $$\dfrac{{{V}_{A}}}{{{V}_{B}}}=\dfrac{{{C}_{B}}}{{{C}_{A}}}\Rightarrow \,\,\,{{V}_{A}}=\dfrac{{{V}_{B}}\times {{C}_{B}}}{{{C}_{A}}}\,\,$$

$$\Rightarrow \,{{V}_{A}}=\dfrac{4\times 3}{2}=6V$$

Potential difference on capacitor A & B, $${{V}_{A}}=6V,\,\,\,\,{{V}_{B}}=4V$$   

Initially wire between both capacitors is neutral, as the source is connected. Positive charge moves toward B and negative charge move toward A.

when the dielectric is placed between the capacitor plate charge increase, due to which wire ends B having more positive charge flow toward it and more positive charge flow toward B.

Hence, charge on capacitor $${{C}_{1}}$$ also increase

Given,

Capacitance, $${{C}_{1}}=0.1\,\mu F\,\,and\,\,{{C}_{2}}=1\,\mu F$$

In series charge is equal

$$ Q={{C}_{1}}{{V}_{1}}={{C}_{2}}{{V}_{2}} $$

$$ {{V}_{2}}=\dfrac{{{C}_{1}}{{V}_{1}}}{{{C}_{2}}} $$

In series total potential difference is sum of all paternal difference

$$ V={{V}_{1}}+{{V}_{2}} $$

$$ V={{V}_{1}}+\dfrac{{{C}_{1}}{{V}_{1}}}{{{C}_{2}}}={{V}_{1}}\left( \dfrac{{{C}_{2}}+{{C}_{1}}}{{{C}_{2}}} \right) $$

$$ {{V}_{1}}=\dfrac{{{C}_{2}}V}{{{C}_{2}}+{{C}_{1}}}=\dfrac{1\times 500}{1+0.1}=454.54\,V $$

Hence, Potential difference across $$0.1\,\mu F\,\,\,is\,\,\,454.5\,V$$ 

The rate of growth of charge for the capacitor

$$q=\varepsilon C\left( 1-{{e}^{\dfrac{-t}{RC}}} \right)$$

Let E be the energy stored inside the capacitor.

Then,

  $$ E=\dfrac{{{q}^{2}}}{2C} $$

 $$ E=\dfrac{{{\varepsilon }^{2}}{{C}^{2}}}{2C}{{\left( 1-{{e}^{\dfrac{-t}{RC}}} \right)}^{2}} $$

 $$ E=\dfrac{{{\varepsilon }^{2}}C}{2}\left( 1-{{e}^{\dfrac{-t}{RC}}} \right) $$

Let r be the rate of energy stored inside the capacitor.

Then,

  $$ r=\dfrac{dE}{dt} $$

 $$ r=\dfrac{2{{\varepsilon }^{2}}C}{2}\left( 1-{{e}^{\dfrac{-t}{RC}}} \right)\left( -{{e}^{\dfrac{-t}{RC}}} \right)\left( \frac{-1}{RC} \right) $$

 $$ r=\dfrac{{{\varepsilon }^{2}}}{R}\left( 1-{{e}^{\dfrac{-t}{RC}}} \right)\left( {{e}^{\dfrac{-t}{RC}}} \right) $$

Now, at $$t = 0$$

The rate of energy stored is

$$r=\dfrac{{{\varepsilon }^{2}}}{R}$$

Hence, the rate of energy stored inside the capacitor is $$\dfrac{{{\varepsilon }^{2}}}{R}$$ 

$${V_1} = {Q \over {{C_1}}},{V_2} = {Q \over {{C_2}}}$$

$$ - {\varepsilon _1} + {V_1} - {\varepsilon _2} + {V_2} = 0$$

$$\left( {{\varepsilon _1} + {\varepsilon _2}} \right) = {V_1} + {V_2}$$

$${\varepsilon _1} + {\varepsilon _2} = {Q \over {{C_1}}} + {Q \over {{C_2}}}$$

$$Q = \left( {{\varepsilon _1} + {\varepsilon _2}} \right)\left( {{{{C_1}{C_2}} \over {{C_1} + {C_2}}}} \right)$$

$${V_2} = {Q \over {{C_2}}}$$

$${V_2} = \left( {{\varepsilon _1} + {\varepsilon _2}} \right)\left( {{{{C_1}} \over {{C_1} + {C_2}}}} \right)$$