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This is one of our many quizzes on Capacitors. Click on this to access more such competitive quizzes on Electric Potential and Capacitance.

Capacitors Quiz 2

physics

Find the ratio of potential difference between $AB$ and $XY$. <img src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/c7a8d7b2-5c49-48cb-b7e5-b3ccdabe2b152848733651030450426.png"/>

 In a parallel combination of capacitors, voltage is equal across each branch, while for capacitors in series, the voltage divides in inverse ratio of the capacitance. 

  $\text{Let the potential difference across the capacitor between AB be V'.} \\ \text{Thus the potential difference across the three '2C' capacitors will be V' too as they are in parallel to }{{C}_{AB}}. \\ \text{ As all }'2C' \text{capacitors are equal, and in series voltage across capacitor }{{C}_{XY}}\text{ will be }\dfrac{V'}{3}. \\ \text{Because in series,voltage divides in inverse ratio of capacitances and all the capacitors have '2C' capacitance}\text{.} \\ \text{Thus }\dfrac{{{C}_{AB}}}{{{C}_{XY}}}=\dfrac{V'}{\dfrac{V'}{3}}=\dfrac{3}{1} \\ \therefore {{C}_{AB}}:{{C}_{XY}}=3:1$

If area of parallel plate capacitor is increased by $20\%$ and distance between its plates is decreased by $20\%$ then capacitance will

 Capacitance of a parallel plate capacitor is $C=\dfrac{{{\varepsilon }_{o}}A}{d}$, where $C$ is capacitance, $A$ is area of plates and $d$ is the separation between the plates. 

  $\text{For a parallel plate capacitor,} \\ C=\dfrac{{{\varepsilon }_{o}}A}{d} \\ \text{Given, the area is increased to }20\%\text{ and separation is decreased by }20\% \\ \text{Thus }A'=1.2\text{ }A,\text{ }d'=0.8d \\ \text{The new capacitance }C'\text{ will be} \\ C'=\dfrac{{{\varepsilon }_{o}}A}{d}=\dfrac{{{\varepsilon }_{o}}\times 1.2A}{0.8d}=\dfrac{3}{2}\dfrac{{{\varepsilon }_{o}}A}{d} \\ C'=1.5C \\ \Rightarrow \Delta C=C'-C=1.5C-C=0.5C \\ \Rightarrow \%\text{ increase in capacitance is }=\dfrac{\Delta C}{C}\times 100=\dfrac{0.5C}{C}\times 100 \\ \therefore \%\text{ increase in capacitance is }=50\%$

In the given circuit, the potential difference across $1\mu \,F$ capacitor is <img src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/5f57b869-088a-4010-9b5b-7da63b18df482444396895494486074.png"/>

 Solve the circuit and make it equivalent to two capacitors in series, then ${{V}_{1}}=\left( \dfrac{{{C}_{2}}}{{{C}_{1}}+{{C}_{2}}} \right)V$ for series combination of ${{C}_{1}}$, ${{C}_{2}}$ capacitor. 

<img src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/4dd93a78-c6f6-4796-8d1c-7df06accd35c267867587510948653.png"/> <img src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/66f87ee4-3396-4bcf-bcc3-fbe7404cf17b1342486878203817761.png"/> <img src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/9b4c7318-0561-4997-8ee7-cb14d7ad25065082577770652825278.png"/> $\text{In step }(1)\text{, }2\text{ }\mu F\text{ capacitors between }A\text{ and }B\text{ are in series}\text{. Their }equivalent\text{ capacitance} \\ {{\text{C}}_{AB}}=\dfrac{C\times C}{C+C} \\ \Rightarrow {{\text{C}}_{AB}}=\dfrac{2\times 2}{2+2} \\ \Rightarrow {{\text{C}}_{AB}}=1\text{ }\mu F \\ \text{The }1\text{ }\mu F\text{ capacitors in step }(2)\text{ are in parallel}\text{. Their equivalent capacitance is} \\ {{\text{C}}_{P}}=C'+C' \\ \Rightarrow {{\text{C}}_{P}}=1+1 \\ \Rightarrow {{\text{C}}_{P}}=2\text{ }\mu F \\ \text{Using voltage divider rule for capacitor }{{C}_{2}} \\ \Rightarrow {{V}_{2}}=\left( \dfrac{{{C}_{1}}}{{{C}_{1}}+{{C}_{2}}} \right){{V}_{total}} \\ \Rightarrow {{V}_{2}}=\dfrac{2}{2+2}\times 10 \\ \therefore {{V}_{2}}=5\text{ }V$

The plates of a capacitor have an area of $10\,{{m}^{2}}$ each. The capacitor in connected to a voltage source of $1000\,V$ and the plates are started to be pushed away from each other at a rate of ${{10}^{-3}}\,m{{s}^{-1}}$. Find the magnitude current through the circuit when the plate separation reaches $1\,cm$. 

 Use $I=\dfrac{dQ}{dt}$ and for a capacitor, $Q=CV$. 

 $\text{As }Q=CV \\ \text{And }C=\dfrac{{{\varepsilon }_{0}}A}{D},D\text{ }is\text{ }plate\text{ }separation \\ \therefore Q=\dfrac{{{\varepsilon }_{0}}A}{D}V \\ \text{Differentiating with respect to time} \\ \Rightarrow \dfrac{dQ}{dt}=-\dfrac{{{\varepsilon }_{0}}A}{{{D}^{2}}}V\dfrac{dD}{dt} \\ \Rightarrow I=-\dfrac{{{\varepsilon }_{0}}AV}{{{D}^{2}}}\dfrac{dD}{dt} \\ \text{As }\dfrac{dD}{dt}={{10}^{-3}}m{{s}^{-1}},D=1cm={{10}^{-2}}m \\ \Rightarrow V=1000V,A=10{{m}^{2}} \\ \Rightarrow \left| I \right|=\dfrac{8.85\times {{10}^{-12}}\times 10\times 1000}{{{\left( {{10}^{-2}} \right)}^{2}}}\times {{10}^{-3}} \\ \Rightarrow \left| I \right|=8.85\times {{10}^{-12-3+8}} \\ \therefore \left| I \right|=8.85\times {{10}^{-7}}\,A$

A capacitor filled with a dielectric constant $10$ has a capacitance of $2\,F$. It is connected to a battery of $10\text{ V}$ and charged complexly. One fully charged, the battery is disconnected and the dielectric slab is taken out from the space between the plate’s. The work done by the capacitor in the process is  

 The work done by the capacitor is work done by its electric field which is conservative in nature.For any conservative force, ${{W}_{conservative}}=-\Delta U$, $U$ is potential energy of the system. 

 $\text{Initially }C=2\text{ F, }V=10V \\ \text{As }Q=CV=2\times 10=20\text{ }C \\ \Rightarrow Q=20\text{ }C \\ \text{When the dielectric is taken out, the capacitance becomes} \\ \Rightarrow C'=\dfrac{C}{k}=\dfrac{2}{10}=\dfrac{1}{5}F \\ \text{As the battery has been disconnected, the capacitor is isolate and charge on the capacitor remains} \\ \text{conserved}\text{.} \\ \text{The initial potential energy }{{U}_{i}}=\dfrac{1}{2}\dfrac{{{Q}^{2}}}{C} \\ \Rightarrow {{U}_{i}}=\dfrac{1}{2}\dfrac{{{\left( 20 \right)}^{2}}}{2}=100\text{ }J \\ \text{The final potential energy of capacitor }{{U}_{t}}=\dfrac{1}{2}\dfrac{{{Q}^{2}}}{C'} \\ \Rightarrow \dfrac{1}{2}\dfrac{{{\left( 20 \right)}^{2}}}{\dfrac{1}{5}}=1000\text{ J} \\ \text{Change in potential energy }\Delta U={{U}_{f}}-{{U}_{i}} \\ \Rightarrow \Delta U=1000-100 \\ \therefore \Delta U=900\text{ J}$