A parallel plate capacitor C with plates of unit area and separation d is filled with a liquid of dielectric constant \$K = 2.\$The level of liquid is \$\dfrac{d}{3}\$ initially. Suppose the liquid level decreases at a constant speed V, the time constant as a function of time \$t\$ is :

At time t the level of liquid is, \$ \dfrac { d }{ 3 }\$\$ -Vt\\ { C }_{ eq }=\dfrac { \dfrac { { \epsilon }_{ 0 }KA }{ \dfrac { d }{ 3 } -Vt } \cdot \dfrac { { \epsilon }_{ 0 }A }{ \dfrac { 2d }{ 3 } +Vt } }{ \dfrac { { \epsilon }_{ 0 }KA }{ \dfrac { d }{ 3 } -Vt } +\dfrac { { \epsilon }_{ 0 }A }{ \dfrac { 2d }{ 3 } +Vt } } =\dfrac { { \epsilon }_{ 0 }KA }{ K\left( \dfrac { 2d }{ 3 } +Vt \right) +\dfrac { d }{ 3 } -Vt } \\ =\dfrac { 2{ \epsilon }_{ 0 } }{ \left( \dfrac { 4d }{ 3 } +2Vt \right) +\dfrac { d }{ 3 } -Vt } =\dfrac { 2{ \epsilon }_{ 0 } }{ \left( \dfrac { 5d }{ 3 } +Vt \right) } =\dfrac { 6{ \epsilon }_{ 0 } }{ 5d+3Vt } \\ \tau =R{ C }_{ eq }=\dfrac { 6{ \epsilon }_{ 0 }R }{ 5d+3Vt } \$

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>>Class 12
>>Physics
>>Electrostatic Potential and Capacitance
>> $Hard Questions$

Electrostatic Potential And Capacitance

Q: A capacitor is filled with an insulator and a certain potential difference is applied to its plates. The energy stored in the capacitor is \$U\$. Now the capacitor is disconnected from the source and the insulator is pulled out of the capacitor. The work performed against the forces of the electric field in pulling out the insulator is \$4U\$. Then the dielectric constant of the insulator is :

Correct answer: Option C Hint: A capacitor is defined as an electrical component which stores energy electrostatically in an electric field. Step 1: Finding final energy The capacitor originally had the energy U and the work required to pull the insulator out is 4U so the final energy is equal to 5U. \$W + U = {U_t}\$ \${U_t} = 4U + U\$ \${U_t} = 5U\$ Step 2: Stored energy in terms of capacitance The capacitance of the insulator with an insulator, \$C = \dfrac{{A{\varepsilon _0}}}{d}k\$ The formula of the energy stored in the insulator is given by, \$U = \dfrac{{{q^2}}}{{2C}}\$ Where capacitance is C the area of the insulator is A the dielectric constant is k the distance between the plates is d and k is permittivity of dielectric constant. When we remove the insulator from the capacitor then the capacitance gets \${C_0}\$\$U=\dfrac{{{q^2}}}{{2C}}\$ \${C_0} = \dfrac{{A{\varepsilon _0}}}{d}\$ And the stored energy will be, \${U_t} = \dfrac{{{q^2}}}{{2{C_0}}}\$ Since, \${U_t} = 5U\$ Step 3: Substituting values Replacing the value of \$U\$ Ut=5U and \${U_t}\$ Uin the above equation we get, \$(\dfrac{{{q^2}}}{{2{C_0}}}) = 5(\dfrac{{{q^2}}}{{2C}})\$ \$\dfrac{C}{{{C_0}}} = 5\$ Replacing the value of \$C\$ and \${C_0}\$ in the above relation \$\dfrac{{\dfrac{{A{\varepsilon _0}}}{d}k}}{{\dfrac{{A{\varepsilon _0}}}{d}}} = 5\$ \$k = 5\$ The dielectric constant value is equal to 5

Q: ‘A’ and ‘B’ are two condensers of capacities 2 \$\mu F \$ and 4\$\mu F\$ They are charged to potential differences of 12V and 6V respectively. If they are now connected (+ve to +ve), the charge that flows through the connecting wire is :

\$Q_A=C_A\ V_A\$\$=2\times 10^{-6}\times 12\$\$Q_A=24\times 10^{-6}\$\$Q_B=C_B\ V_B\$\$=4\times 10^{-6}\times 6\$\$Q_B=24\times 10^{-6}\$Now when plates are attached total charge \$=Q_A+Q_B\$\$Q=48\times 10^{-6}C\$Let x he charge on A then Q-x charge is on B\$ \Rightarrow x=2\mu F\times V\ and\ Q-x=4\mu \ F\times V.\$\$ \Rightarrow \frac{x}{Q-x}=\frac{1}{2}\ \ \ 2x=Q-x\ \ \ \ x=\frac{9}{3}\$\$x=16\times 10^{-6};Q-x=32\times 10^{-6}\$Now charge flows from A to B is given by,\$=final -initial\$\$=32\times 10^{-6}-20\times 10^{-6}\$\$ \bigtriangleup Q=8\times 10^{-6} C\$\$\bigtriangleup Q = 8\mu C\$

Q: Two identical capacitors are connected as shown in the figure. A dielectric slab is introduced between the plates of one of the capacitors so as to fill the gap, the battery remaining connected. The charge on each capacitor will be :(charge on each condenser is \$q_{0}\$; \$k\$ = dielectric constant )

When capacitors are connected in series the charge present on them must be equal.\$\dfrac{Q}{c_{1}}+\dfrac{Q}{c_{2}}=2\$ ------------ (1)Now when \$k\$ is introduced \$c_{2}\$ becomes \$kc_{2}\$Let \$q_{o}\$ be the charge on capacitorgiven \$c_{1}=c_{2}\$\$\Rightarrow 2\dfrac { { q }_{ o } }{ c } =2,\dfrac { { q }_{ o } }{ c } =1\$ from (1) \$ c ={ q }_{ o } \$After introducing the dielectric, let charge on each plate become \$Q_{ 1 }\$\$\Rightarrow \dfrac { Q_{ 1 } }{ { q }_{ o } } +\dfrac { Q_{ 1 } }{ k{ q }_{ 0 } } =2\$\$\Rightarrow \dfrac { kQ_{ 1 }+Q_{ 1 } }{ k{ q }_{ o } } =2\$\$\Rightarrow Q_{ 1 }=\dfrac { 2kq_{ o } }{ k+1 } \$\$\Rightarrow Q_{1}=\dfrac{2q_{o}}{1+ \dfrac{1}{k}}\$

Q: A dielectric slab of thickness \$d\$ is inserted in a parallel plate capacitor whose negative plate is at \$x = 3d\$. The slab is equidistant from the plates. The capacitor is given some charge. As \$x\$ goes from \$0\$ to \$3d\$ :

As \$x\$ from \$0-3d\$1) Magnitude of \$\overset { \rightarrow }{ E } \$ changes due to dielectric slab. In region A and C is same but not in B2) the direction of \$\overset { \rightarrow }{ E } \$ remains same throughout in all regions A,B,C3) Electric potential also changes, since electric field varies, so potential also changes.4) Electric potential first increases and then decreases due to dielectric slab and then again increases.

Q: The capacity between the points A and B in the given circuit will be:

The charge distribution is as shown in the figure.Using Kirchhoff's law,\$ V - q1/C_1 + (q2-q1)/C_3 - q1/C_1 =0 \Rightarrow V = 2q1/C_1 - (q2-q1)/C_3 \$\$ V - q2/C_2 + (q1-q2)/C_3 - q2/C_2 =0 \Rightarrow V = 2q2/C_2 - (q1-q2)/C_3 \$Therefore,\$ C_1C_3 V = 2C_3q1 - C_2 q2 + C_2q1 = (2C_3+C_1)q1 - C_1q2\$\$ C_2C_3 V = 2C_3q2 - C_2q1 + C_2q2 = (2C_3 + C_2)q2 - C_2q1\$ Solving these two equations we get,\$q1 = \dfrac{C_1(C_2+C_3)V}{C_1+C_2+2C_3} \$\$q2 = \dfrac{C_2(C_1 + C_3)V}{C_1+C_2+2C_3} \$Equivalent capacitance \$=Q_{net}/V = (q1+q2)/V = \dfrac{2C_1C_2 + C_1C_3 + C_2C_3}{C_1 +C_2 + 2C_3}\$

Q: In the network of capacitors shown in the diagram, the terminal A is at a potential of \$20 V\$ and the terminal B is grounded. If \$1 F\$ capacitor is charged to \$10C\$, then :

Equivalent capacitance between AB is given by \$\dfrac{1}{C_{AB}} = \dfrac{1}{\dfrac{1}{\dfrac{1}{3} + \dfrac{1}{3} + \dfrac{1}{3}} +2} + \dfrac{1}{1 + \dfrac{1}{\dfrac{1}{4} + \dfrac{1}{4}}}\$ \$\dfrac{1}{C_{AB}} = \dfrac{1}{3} + \dfrac{1}{3} = \dfrac{2}{3}\$ Hence , equivalent capacitance \$= 1.5 F\$ Charge on 1F\$=10C\Rightarrow V_{c}=V_{D}=\dfrac{10}{1}=10V\$ Since\$V_{A}=20V\$, Potential drop across the \$2 F\$ capacitor , \$V_{A}-V_{C}=20-10=10V\$ Equivalent capacitance of the three 3\$F\$ capacitances in series is \$=\dfrac{1}{C}=\dfrac{1}{3}+\dfrac{1}{3}+\dfrac{1}{3}\$ \$\therefore C=1F\$ \$\therefore \$ charge on each 3F\$=10\times 1=10C\$ Further, charge on +ve plate of \$1.5 F\$ capacitor\$=+30C\$ \$\therefore \$ charge leaking to the ground (to make second plate \$-30C\$) is \$+30C\$

Q: The capacitance \$C_{AB}\$ in the given network is:

From the given circuit, we get \$\dfrac{C_1}{C_2} = \dfrac{5}{10} = \dfrac{1}{2}\$And \$\dfrac{C_3}{C_4} = \dfrac{10}{5} = \dfrac{2}{1}\$Thus \$\dfrac{C_1}{C_2}\$ is reciprocal of \$\dfrac{C_3}{C_4}\$For this particular type of circuit, the equivalent capacitance is given by-\$C_{net} = \dfrac{2C_1C_2 + C_3(C_1+C_2)}{C_1+C_2+2C_3} \$\$C_{net} = \dfrac{2(5)(10) + 5(5+10)}{5+10+2(5)} = \dfrac{175}{25} = 7\mu F\$

Q: The equivalent capacitance \$C_{AB}\$ of the circuit shown in the figure is :

The given circuit can be reconstructed as shown (As the potential difference between A and B has to same through all possible paths) Using the laws of combination of capacitors, \$ C_{net} = C_1 + C_2 + C_3 \,where\,\$ \$ C_2 = \dfrac{C}{2} , \dfrac{1}{C_1} = \dfrac{1}{C} + \dfrac{1}{C} + \dfrac{1}{(C+ \dfrac{C}{2})} = \dfrac{2}{C} + \dfrac{2}{3C} = \dfrac{8}{3C} \Rightarrow C_1 = \dfrac{3C}{8} \$ Similarly, \$C_3 = \dfrac{3C}{8} \$ \$C_{net} = \dfrac{6C}{8} + \dfrac{C}{2} = \dfrac{5C}{4} \$

Physics
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Easy Qs Med Qs Hard Qs

A capacitor is filled with an insulator and a certain potential difference is applied to its plates. The energy stored in the capacitor is $U$ . Now the capacitor is disconnected from the source and the insulator is pulled out of the capacitor. The work performed against the forces of the electric field in pulling out the insulator is $4 U$ . Then the dielectric constant of the insulator is :

4

8

5

3

Hard

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‘A’ and ‘B’ are two condensers of capacities 2 $μ F$ and 4 $μ F$ They are charged to potential differences of 12V and 6V respectively. If they are now connected (+ve to +ve), the charge that flows through the connecting wire is :

μ

C from A to B

μ

C from A to B

μ

C from B to A

μ

C from B to A

Hard

View solution>

Two identical capacitors are connected as shown in the figure. A dielectric slab is introduced between the plates of one of the capacitors so as to fill the gap, the battery remaining connected. The charge on each capacitor will be :(charge on each condenser is $q_{0}$ ; $k$ = dielectric constant )

\frac{2 q _{0}}{1 + 1 / k}

\frac{q _{0}}{1 + 1 / k}

\frac{2 q _{0}}{1 + k}

\frac{q _{0}}{1 + k}

Hard

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A parallel plate capacitor C with plates of unit area and separation d is filled with a liquid of dielectric constant $K = 2 .$ The level of liquid is $\frac{d}{3}$ initially. Suppose the liquid level decreases at a constant speed V, the time constant as a function of time $t$ is :

\frac{6 ϵ _{0} R}{5 d + 3 V t}

\frac{( 1 5 d + 9 V t ) ϵ _{0} R}{2 d ^{2} - 3 d V t - 9 V ^{2} t ^{2}}

\frac{6 ϵ _{0} R}{5 d - 3 V t}

\frac{( 1 5 d - 9 V t ) ϵ _{0} R}{2 d ^{2} + 3 d V t - 9 V ^{2} t ^{2}}

Hard

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A dielectric slab of thickness $d$ is inserted in a parallel plate capacitor whose negative plate is at $x = 3 d$ . The slab is equidistant from the plates. The capacitor is given some charge. As $x$ goes from $0$ to $3 d$ :

the magnitude of the electric field remains the same

the direction of the electric field remains the same

the electric potential increases continuously

the electric potential increases at first, then decreases and again increases

Hard

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A parallel plate capacitor of capacity $5 μ F$ and plate separation $6 c m$ is connected to a $1 V$ battery and is charged. A dielectric of dielectric constant $4$ and thickness $4 c m$ is introduced into the capacitor. The additional charge that flows into the capacitor from the battery is:

μ

μ

μ

μ

Hard

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The capacity between the points A and B in the given circuit will be:

\frac{2 C _{1} C _{2} + C _{3} ( C _{1} + C _{2} )}{C _{1} + C _{2} + 2 C _{3}}

\frac{C _{1} C _{2} + C _{2} C _{3} + C _{3} C _{1}}{C _{1} + C _{2} + C _{3}}

\frac{C _{1} ( C _{2} + C _{3} ) + C _{2} ( C _{1} + C _{3} )}{C _{1} + C _{2} + 3 C _{3}}

\frac{C _{1} C _{2} C _{3}}{C _{1} C _{2} + C _{2} C _{3} + C _{3} C _{1}}

Hard

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In the network of capacitors shown in the diagram, the terminal A is at a potential of $20 V$ and the terminal B is grounded. If $1 F$ capacitor is charged to $10 C$ , then :

the effective capacitance between A and B is

1.5 F

the potential drop across

2 F

capacitor is

20 V

each

3 F

capacitor is charged to

10 C

net charge leaked to ground is zero.

Hard

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The capacitance $C_{A B}$ in the given network is:

7 μ F

\frac{5 0}{7} μ F

7.5 μ F

\frac{7}{5 0} μ F

Hard

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The equivalent capacitance $C_{A B}$ of the circuit shown in the figure is :

\frac{5}{4} C

\frac{4}{5} C

2 C

C

Hard

View solution>

Electrostatic Potential And Capacitance

Physics 10886 Views

‘A’ and ‘B’ are two condensers of capacities 2 μF and 4μF They are charged to potential differences of 12V and 6V respectively. If they are now connected (+ve to +ve), the charge that flows through the connecting wire is :

Two identical capacitors are connected as shown in the figure. A dielectric slab is introduced between the plates of one of the capacitors so as to fill the gap, the battery remaining connected. The charge on each capacitor will be :(charge on each condenser is q0​; k = dielectric constant )

A parallel plate capacitor C with plates of unit area and separation d is filled with a liquid of dielectric constant K=2.The level of liquid is 3d​ initially. Suppose the liquid level decreases at a constant speed V, the time constant as a function of time t is :

A dielectric slab of thickness d is inserted in a parallel plate capacitor whose negative plate is at x=3d. The slab is equidistant from the plates. The capacitor is given some charge. As x goes from 0 to 3d :

A parallel plate capacitor of capacity 5μF and plate separation 6cm is connected to a 1V battery and is charged. A dielectric of dielectric constant 4 and thickness 4cm is introduced into the capacitor. The additional charge that flows into the capacitor from the battery is:

The capacity between the points A and B in the given circuit will be:

In the network of capacitors shown in the diagram, the terminal A is at a potential of 20V and the terminal B is grounded. If 1F capacitor is charged to 10C, then :

The capacitance CAB​ in the given network is:

The equivalent capacitance CAB​ of the circuit shown in the figure is :

Physics
10886 Views

‘A’ and ‘B’ are two condensers of capacities 2 $μ F$ and 4 $μ F$ They are charged to potential differences of 12V and 6V respectively. If they are now connected (+ve to +ve), the charge that flows through the connecting wire is :

Two identical capacitors are connected as shown in the figure. A dielectric slab is introduced between the plates of one of the capacitors so as to fill the gap, the battery remaining connected. The charge on each capacitor will be :(charge on each condenser is $q_{0}$ ; $k$ = dielectric constant )

A parallel plate capacitor C with plates of unit area and separation d is filled with a liquid of dielectric constant $K = 2 .$ The level of liquid is $\frac{d}{3}$ initially. Suppose the liquid level decreases at a constant speed V, the time constant as a function of time $t$ is :

A dielectric slab of thickness $d$ is inserted in a parallel plate capacitor whose negative plate is at $x = 3 d$ . The slab is equidistant from the plates. The capacitor is given some charge. As $x$ goes from $0$ to $3 d$ :

A parallel plate capacitor of capacity $5 μ F$ and plate separation $6 c m$ is connected to a $1 V$ battery and is charged. A dielectric of dielectric constant $4$ and thickness $4 c m$ is introduced into the capacitor. The additional charge that flows into the capacitor from the battery is:

In the network of capacitors shown in the diagram, the terminal A is at a potential of $20 V$ and the terminal B is grounded. If $1 F$ capacitor is charged to $10 C$ , then :

The capacitance $C_{A B}$ in the given network is:

The equivalent capacitance $C_{A B}$ of the circuit shown in the figure is :