Q: Energy stored in a capacitor and dissipated during charging a capacitor bear a ratio:

The energy stored in a capacitor is given by:\$U_f = \dfrac{1}{2} CV^2\$Final energy of capacitor will be \$\dfrac{1}{2}CV^2\$.The work done by the cell is:\$W = q \times\$ (emf of cell)\$= CV \times (V)\$\$=CV^2\$The heat produced is given bt \$\to \$ (work done) \$-\$ (U_f - V_i)\$\$\$H = CV^2 - \left(\dfrac{1}{2} CV^2 - o\right)\$\$H = CV^2 - \dfrac{1}{2}CV^2\$\$= \dfrac{1}{2}CV^2\$So, \$\dfrac{\text{Energy stored}}{\text{Heat produced}} = \dfrac{\dfrac{1}{2}CV^2}{\dfrac{1}{2}CV^2}\$\$= 1:1\$

Q: A capacitor of capacitance \$10 \mu F\$ is charged to \$10 V\$. The energy stored in it is

Given : \$C = 10\mu F = 10^{-5} F\$ \$V =10\$ voltsEnergy stored \$E = \dfrac{1}{2}CV^2\$\$\therefore\$ \$E = \dfrac{1}{2}\times 10^{-5} \times 10^2 = 5\times 10^{-4}\$ \$J\$\$\implies\$ \$E = 500\mu J\$

Q: The amount of work done in increasing the voltage across the plates of capacitor from 5V to 10V is 'W'. The work done in increasing it from 10V to 15V will be

Let the capacitance of the capacitor be \$C\$.Energy stored in the capacitor \$E = \dfrac{1}{2}CV^2\$Work done in changing the potential from 5V to 10V, \$W = \dfrac{1}{2}C (V_2^2 - V_1^2)\$\$\therefore\$ \$W = \dfrac{1}{2}C (10^2 - 5^2) = \dfrac{1}{2}C (75)\$Work done in changing the potential from 10V to 15V, \$W' = \dfrac{1}{2}C (15^2 - 10^2)\$\$\therefore\$ \$W' = \dfrac{1}{2}C (125)\$\$\implies\$ \$\dfrac{W'}{W} = \dfrac{125}{75} = 1.67\$We get \$W' = 1.67 W\$

Q: A 12pF capacitor is connected to a 50V battery. How much electrostatic energy is stored in the capacitor?

The total energy of capacitor is \$E=\dfrac{1}{2}CV^{2}=\dfrac{1}{2} \times 12 \times 10^{-12} \times 50^{2}=1.5 \times 10^{-8}J\$

Q: A 900 pF capacitor is charged by 100 V battery as shown in figure. The capacitor is disconnected from the battery and connected to another 900 pF capacitor. What is the electrostatic energy stored by the system?

We have the initial energy as \$\dfrac{1}{2}CV^2=\dfrac{1}{2}\times 9\times 10^{-10}\times 100^2=4.5\times 10^{-6} J\$Now the capacitor is disconnected and is connected to another 900 pF capacitor. In the steady situation, the two capacitors have their positive plates at the same potential. Let the common potential difference be \$V’\$. The charge on each capacitor is \$Q’ = CV’\$.By charge conservation, \$Q’ = Q/2\$. This implies \$V’ = V/2\$. The total energy of the system is = \$2 × (1/2) Q’V’ = (1/4) QV = 2.25 × 10^{-6} J\$ Thus we get the electrostatic energy stored by the system as \$\dfrac{1}{2}(C_1+C_2)V^2=\dfrac{1}{2}(2\times 9\times 10^{-10})\times 50^2=2.25\times 10^{-6}\$

Q: A parallel plate capacitor of capacitance C is charged to a potential V. It is then connected to another uncharged capacitor having the same capacitance. Find out the ratio of the energy stored in the combined system to that stored initially in the single capacitor.

Energy stored initially, \$U_i = \dfrac{1}{2}CV^2\$After connecting to the other uncharged capacitor, charge is distributed equally.Since, \$Q=CV\$, \$V_f = V/2\$ Hence, energy stored finally in the two capacitors is given by:\$U_f = 2 \times \dfrac{1}{2}CV_f^2\$\$U_f = \dfrac{CV^2}{4}\$Hence, ratio of energy stored in combined system to that stored initially is:\$\dfrac{U_f}{U_i}=\dfrac{1}{2}\$

Question 1

Derive an expression for energy stored in a capacitor. In which form energy is stored?

Accepted Answer

Let us consider a capacitor of capacitance C and potential difference V between the plates.Let the charge on one plate be +q and -q on the other.Suppose the capacitor is being charged gradually.Now,at any stage the charge on capacitor is q.Therefore, the potential difference = $ \frac{q}{C} $Small amount of work doe in giving n additional charge dq to the capacitor isdW= $\dfrac{q}{C}*dq$Total work done in giving a charge Q to the capacitor  isW=$\int dW$W= $ \int_{0}^{Q} \frac{Q}{C}dq $ W = $\dfrac{Q^{2}}{C}$Energy = EE= $\dfrac{Q^2}{2C}$ = $\dfrac {CV^2}{2}$ = $\dfrac{QV}{2}$The energy is stored in the form of potential energy.

Question 2

The energy required to charge a parallel plate condenser of plate separation d and plate area of cross-section A such that the uniform electric field between the plates is E, is :

Accepted Answer

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Question 3

Energy stored in a capacitor and dissipated during charging a capacitor bear a ratio:

Accepted Answer

The energy stored in a capacitor is given by:$U_f = \dfrac{1}{2} CV^2$Final energy of capacitor will be $\dfrac{1}{2}CV^2$.The work done by the cell is:$W = q \times$ (emf of cell)$= CV \times (V)$$=CV^2$The heat produced is given bt $\to $ (work done) $-$ (U_f - V_i)$H = CV^2 - \left(\dfrac{1}{2} CV^2 - o\right)$$H = CV^2 - \dfrac{1}{2}CV^2$$= \dfrac{1}{2}CV^2$So, $\dfrac{\text{Energy stored}}{\text{Heat produced}} = \dfrac{\dfrac{1}{2}CV^2}{\dfrac{1}{2}CV^2}$$= 1:1$

Question 4

A capacitor of capacitance $10 \mu F$ is charged to $10 V$. The energy stored in it is

Accepted Answer

Given :    $C = 10\mu F = 10^{-5} F$                $V =10$ voltsEnergy stored      $E = \dfrac{1}{2}CV^2$$\therefore$   $E = \dfrac{1}{2}\times 10^{-5} \times 10^2 = 5\times 10^{-4}$ $J$$\implies$   $E = 500\mu J$

Question 5

The amount of work done in increasing the voltage across the plates of capacitor from 5V to 10V is 'W'. The work done in increasing it from 10V to 15V will be

Accepted Answer

Let the capacitance of the capacitor be $C$.Energy stored in the capacitor $E = \dfrac{1}{2}CV^2$Work done in changing the potential from 5V to 10V, $W = \dfrac{1}{2}C (V_2^2 - V_1^2)$$\therefore$ $W = \dfrac{1}{2}C (10^2 - 5^2) = \dfrac{1}{2}C (75)$Work done in changing the potential from 10V to 15V, $W' = \dfrac{1}{2}C (15^2 - 10^2)$$\therefore$ $W' = \dfrac{1}{2}C (125)$$\implies$ $\dfrac{W'}{W} = \dfrac{125}{75} = 1.67$We get $W' = 1.67 W$

Question 6

If the charge on a capacitor is increased by $2$ coulomb, the energy stored in it increases by 21%. The original charge on the capacitor is :

Accepted Answer

According to questionPercentage increase in energy of capacitor=$\dfrac { \dfrac { { q }_{ f }^{ 2 } }{ 2C } -\dfrac { { q }_{ i }^{ 2 } }{ 2C }  }{ \dfrac { { q }_{ i }^{ 2 } }{ 2C }  } \times 100=21$ ... (i) and ${ q }_{ f }-{ q }_{ i }=2$   ...(ii)From (i) and (ii)$ 4(q_i + 1)  = 0.21 (q_i)^2 $On solving:Solving, we get ${ q }_{ i }=20C$

Question 7

A 12pF capacitor is connected to a 50V battery. How much electrostatic energy is stored in the capacitor?

Accepted Answer

The total energy of capacitor is $E=\dfrac{1}{2}CV^{2}=\dfrac{1}{2} \times 12 \times 10^{-12} \times 50^{2}=1.5 \times 10^{-8}J$

Question 8

A 900 pF capacitor is charged by 100 V battery as shown in figure. The capacitor is disconnected from the battery and connected to another 900 pF capacitor. What is the electrostatic energy stored by the system?

Accepted Answer

We have the initial energy as $\dfrac{1}{2}CV^2=\dfrac{1}{2}\times 9\times 10^{-10}\times 100^2=4.5\times 10^{-6}  J$Now the capacitor is disconnected and is connected to another 900 pF capacitor. In the steady situation, the two capacitors have their positive plates at the same potential. Let&#10;the common potential difference be $V&#8217;$. The charge on each capacitor is $Q&#8217; = CV&#8217;$.By charge&#10;conservation, $Q&#8217; = Q/2$. This implies $V&#8217; = V/2$. The total energy of the system is = $2 &#215; (1/2) Q&#8217;V&#8217; =&#10;(1/4) QV = 2.25 &#215; 10^{-6} J$ Thus we get the electrostatic energy stored by the system as $\dfrac{1}{2}(C_1+C_2)V^2=\dfrac{1}{2}(2\times 9\times 10^{-10})\times 50^2=2.25\times 10^{-6}$

Question 9

A parallel plate capacitor of capacitance C is charged to a potential V. It is then connected to another uncharged capacitor having the same capacitance. Find out the ratio of the energy stored in the combined system to that stored initially in the single capacitor.

Accepted Answer

Energy stored initially, $U_i = \dfrac{1}{2}CV^2$After connecting to the other uncharged capacitor, charge is distributed equally.Since, $Q=CV$,            $V_f = V/2$ Hence, energy stored finally in the two capacitors is given by:$U_f = 2 \times \dfrac{1}{2}CV_f^2$$U_f = \dfrac{CV^2}{4}$Hence, ratio of energy stored in combined system to that stored initially is:$\dfrac{U_f}{U_i}=\dfrac{1}{2}$

Question 10

A parallel plate capacitor has a uniform electric field $E$ in the space between the plates. If the distance between the plates is $d$ and area of each plate is $A$, the energy stored in the capacitor is :

Accepted Answer

Hint: The capacitance is defined as the product of permittivity of free space and the area divided by the separation between the plates.&#10;&#10;Step1: finding total&#10;energy&#10;&#10;Total energy $=\dfrac{1}{2}&#10;\mathrm{CV}^{2}=$ energy density $\times$ volume, where $\mathrm{C}=\dfrac{\epsilon_{0}&#10;\mathrm{~A}}{\mathrm{~d}}, \mathrm{~V}=\mathrm{Ed}$ $\mathrm{U}=\left(\dfrac{1}{2}&#10;\epsilon_{0} \mathrm{E}^{2}\right)(\mathrm{Ad})$&#10;&#10;Correct answer:  (B)

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Energy Stored in Capacitor - Mathematically

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Energy stored in the capacitor

Problem-based on Energy of Capacitor - I

Energy Stored in Parallel Plate Capacitor

Problem based on Energy stored in Capacitor

Derive an expression for energy stored in a capacitor. In which form energy is stored?

The energy required to charge a parallel plate condenser of plate separation d and plate area of cross-section A such that the uniform electric field between the plates is E, is :

Energy stored in a capacitor and dissipated during charging a capacitor bear a ratio:

A capacitor of capacitance $10 μ F$ is charged to $10 V$ . The energy stored in it is

The amount of work done in increasing the voltage across the plates of capacitor from 5V to 10V is 'W'. The work done in increasing it from 10V to 15V will be

If the charge on a capacitor is increased by $2$ coulomb, the energy stored in it increases by 21%. The original charge on the capacitor is :

A 12pF capacitor is connected to a 50V battery. How much electrostatic energy is stored in the capacitor?

A 900 pF capacitor is charged by 100 V battery as shown in figure. The capacitor is disconnected from the battery and connected to another 900 pF capacitor. What is the electrostatic energy stored by the system?

A parallel plate capacitor of capacitance C is charged to a potential V. It is then connected to another uncharged capacitor having the same capacitance. Find out the ratio of the energy stored in the combined system to that stored initially in the single capacitor.

A parallel plate capacitor has a uniform electric field $E$ in the space between the plates. If the distance between the plates is $d$ and area of each plate is $A$ , the energy stored in the capacitor is :

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Energy Stored in Capacitor - Mathematically

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Energy stored in the capacitor

Problem-based on Energy of Capacitor - I

Energy Stored in Parallel Plate Capacitor

Problem based on Energy stored in Capacitor

Derive an expression for energy stored in a capacitor. In which form energy is stored?

The energy required to charge a parallel plate condenser of plate separation d and plate area of cross-section A such that the uniform electric field between the plates is E, is :

Energy stored in a capacitor and dissipated during charging a capacitor bear a ratio:

A capacitor of capacitance 10μF is charged to 10V. The energy stored in it is

The amount of work done in increasing the voltage across the plates of capacitor from 5V to 10V is 'W'. The work done in increasing it from 10V to 15V will be

If the charge on a capacitor is increased by 2 coulomb, the energy stored in it increases by 21%. The original charge on the capacitor is :

A 12pF capacitor is connected to a 50V battery. How much electrostatic energy is stored in the capacitor?

A 900 pF capacitor is charged by 100 V battery as shown in figure. The capacitor is disconnected from the battery and connected to another 900 pF capacitor. What is the electrostatic energy stored by the system?

A parallel plate capacitor of capacitance C is charged to a potential V. It is then connected to another uncharged capacitor having the same capacitance. Find out the ratio of the energy stored in the combined system to that stored initially in the single capacitor.

A parallel plate capacitor has a uniform electric field E in the space between the plates. If the distance between the plates is d and area of each plate is A, the energy stored in the capacitor is :

A capacitor of capacitance $10 μ F$ is charged to $10 V$ . The energy stored in it is

If the charge on a capacitor is increased by $2$ coulomb, the energy stored in it increases by 21%. The original charge on the capacitor is :

A parallel plate capacitor has a uniform electric field $E$ in the space between the plates. If the distance between the plates is $d$ and area of each plate is $A$ , the energy stored in the capacitor is :