Electrostatic Potential and Capacitance Test - 55

Wheat stone bridge will be formed

$${{C}_{pq}}={{C}_{ST}}=\dfrac{1}{\dfrac{1}{C}+\dfrac{1}{C}}=\dfrac{C}{2}$$

$${{C}_{AB}}={{C}_{pq}}+{{C}_{ST}}=\dfrac{C}{2}+\dfrac{C}{2}=C$$

Hence, effective capacitance is $$1\,C$$ 

Voltage across $$2\,\mu F$$capacitor is $$\dfrac{24\,}{2}=12\,V$$

So, voltage across $$6\mu F$$ capacitor is $$18-12=6\,V$$

Hence, charge on $$6\,\mu F$$ capacitor$$6\times 6=36\,\mu C$$

So, charge on C capacitor $$=36-24=12\,\mu C$$

The value of C is $$C=\dfrac{12}{12}=1\,\mu F$$

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 $$0.1\,\mu F\,\,\,is\,\,\,454.5\,V$$ 

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