Informative line

Energy Stored In A Capacitor

Learn how to find the total energy stored in each capacitor of a circuit and calculate the work done by battery when the initial charge present on capacitor and the change in potential energy when the polarity of battery is reversed.

Energy Stored in Capacitor 

  • Consider a parallel plate capacitor of area A, having charge "+ Q" on one plate and charge "– Q" on another plate.
  • Electric field due to positive plate of capacitor is given as,

  \(E_+ = \dfrac{\sigma}{2\epsilon _o} = \dfrac{Q}{2A\epsilon_ o}\) 

  • Force on negative plate of capacitor due to field of positive plate is given as,

              \(F=-QE\)

              \(F= -Q.\dfrac{Q}{2A\epsilon _o}\)

              \(F = \dfrac{–Q^{2}}{2A\epsilon_o}\)

  • Magnitude of force is given as, 

                \(F = \dfrac{Q^{2}}{2A\epsilon_o}\) ( Force of attraction )

  • Force of attraction is the force by which both the plates attract each other.
  • Initially these plates are kept quite close to each other, i.e., separation between the plates is very small.
  • Now one plate is kept fixed and other plate is moved away slowly till the separation between the plates increases from zero to "d".
  • Since, there is a force of attraction between plates of capacitor so, work is done to separate them upto distance "d".

           \(W = Fd\)

          \(W = \dfrac{Q^{2}d}{2A\epsilon_0}\)  

  or      \(W = \dfrac{Q^{2}}{2 C}\)    

  • This work  done to separate the plate is stored as the increase of energy in the system which is same as the amount of energy stored in capacitor.

            \(U = \dfrac{Q^{2}}{2 C}\)

  • Consider a capacitor of any type to be in charging process.
  • At any instant, the instantaneous value of charge on the plates of capacitor are +q and –q respectively.

 

 

  • The instantaneous value of potential difference is 

               \(\Delta V = \dfrac{q}{C}\)

  • At the next instant of time, \( dq\)  amount of charge is transferred from negative plate to positive plate.
  • Work must be done to transfer \( dq\)  amount of charge from negative plate to positive plate,i.e., to move charge from plate at lower potential to plate at higher potential.

             \(dW=\dfrac{q} {C}dq \)

  •  Since, the capacitor can be charged up to Q so, the total work done in charging is

             \(\int\limits ^W_0dW = \int\limits^Q_0 \dfrac {q}{C} dq \)

or       \(W = \dfrac{Q^{2}}{2 C}\) 

  • This work done in charging the capacitor up to Q appears as energy stored in capacitor.

            \(U = \dfrac{Q^{2}}{2 C}\)

Illustration Questions

Consider a parallel plate capacitor, having charge Q and capacitance C. What will happen if the separation between plates is doubled, given the potential difference across capacitor is \(\Delta \)V? ( Assume capacitor is not connected to battery )  

A Charge is doubled 

B Charge is halved 

C Potential difference is doubled

D Potential difference is halved

×

The capacitance of a capacitor is given as

\(C = \dfrac {\epsilon_0 A}{d}\)

where ,C = Capacitance 

           A = Area of plates 

           d = Separation between plates 

When separation between plates is doubled, 

\(d^{'} = 2 d\)

then, the new capacitance is given as, 

\(C^{'} = \dfrac {\epsilon_0 A}{2d}\)

 or, \(C^{'} = \dfrac {C}{2}\)                   

Now charge on capacitor is given as,

 \(Q=C^{'}V\)

Since, the charge remains constant so,

\(C\propto \dfrac {1}{V}\)

Hence, when capacitance is halved, the potential difference is doubled.  

Consider a parallel plate capacitor, having charge Q and capacitance C. What will happen if the separation between plates is doubled, given the potential difference across capacitor is \(\Delta \)V? ( Assume capacitor is not connected to battery )  

A

Charge is doubled 

.

B

Charge is halved 

C

Potential difference is doubled

D

Potential difference is halved

Option C is Correct

Energy Stored in each Capacitor of a Circuit

  • Consider a circuit having two capacitors in series of capacitance C1 and C2 respectively, are connected with a battery of e.m.f. \(\mathcal{E}\) as shown in figure. 

  • Since capacitors C1 and C2 are connected in series , so charge stored on both the capacitors, is same, i.e., Q.
  • Energy stored in capacitor C1

                     \(U_1 =\dfrac {Q^{2}}{2C_1}\) 

  • Energy stored in capacitor C2

                      \(U_2 =\dfrac {Q^{2}}{2C_2}\)

  • Then, total energy of the system is given as 

            U = U1 + U2

          \(\Rightarrow U = \dfrac {Q^{2}}{2C_1} + \dfrac {Q^{2}}{2 C_2}\)

         \(\Rightarrow U = Q^{2} ( \dfrac {1}{2C_1} + \dfrac{1}{2C_2})\)

        \(\Rightarrow U = \dfrac {Q^{2}}{2} \left (\dfrac {1}{C_1} + \dfrac{1}{C_2} \right)\)

Illustration Questions

Two capacitors of capacitance C1 = 12 \(\mu\)F and C2 = 6 \(\mu\)F, are connected in series with \(\mathcal{E}\)= 3 V battery. Calculate the energy stored in each capacitor.

A  \(8\,\mu J,\,12\,\mu J\)

B \(6\,\mu J,\,12\,\mu J\)

C \(4\,\mu J,\,12\,\,\mu J\)

D \(12\,\mu J,\,3\,\mu J\)

×

Equivalent capacitance when C1 and C2 are connected in series

\(\dfrac {1}{C_{eq}} = \dfrac {1}{C_1} + \dfrac {1}{C_2}\)

or      \(\dfrac {1}{C_{eq}} = \dfrac {1}{12 × 10^{-6}} + \dfrac {1}{6 × 10^{-6}}\)

or    \(\dfrac {1}{C_{eq} } = \dfrac {1}{4 \,\mu F} \)

or, \(C_{eq} = 4\, \mu F\)

 

image

Charge on each capacitor connected in series, is same and given as

 \(Q=C\mathcal{E}\)

or    \(Q = 4 × 10 ^{-6} × 3 \)

or   \( Q = 12 \,\)\(\mu C\)

 

image

Energy stored in capacitor C1

or,   \(U_1 = \dfrac {Q^{2}}{2 C_1}\)

or,   \(U_1 = \dfrac {(12 × 10 ^{-6})^{2}}{2 × (12 × 10 ^{-6})}\)

or,  \(U_1 = 6 \,\mu J\)

 

image

Energy stored in capacitor C2

or,   \(U_2 = \dfrac {Q^{2}}{2 C_2}\)

or,   \(U_2 = \dfrac {(12 × 10 ^{-6})}{2 × 6 × 10 ^{-6}}^{2}\)

or,  \(U_2 = 12 \,\mu J\)

 

image

Two capacitors of capacitance C1 = 12 \(\mu\)F and C2 = 6 \(\mu\)F, are connected in series with \(\mathcal{E}\)= 3 V battery. Calculate the energy stored in each capacitor.

A

 \(8\,\mu J,\,12\,\mu J\)

.

B

\(6\,\mu J,\,12\,\mu J\)

C

\(4\,\mu J,\,12\,\,\mu J\)

D

\(12\,\mu J,\,3\,\mu J\)

Option B is Correct

Total Energy Stored in a Circuit 

  • Consider two capacitors having capacitance C1 and C2 are connected across a battery of e.m.f. \(\mathcal{E}\), as shown in figure.

 

  • Energy stored in capacitor C1

          \(U_1 = \dfrac {1}{2} C_1\mathcal{E} ^{2}\)

  • Energy stored in capacitor C2

        \(U_2 = \dfrac {1}{2} C_2 \mathcal{E} ^{2}\)

  • Total energy of the circuit 

     \(U = U_1 + U_2\)

\(\Rightarrow U = \dfrac{1}{2} C_1 \mathcal{E}^{2} + \dfrac {1}{2} C_2 \mathcal{E}^{2} \)

\(\Rightarrow U = \dfrac{1}{2}\mathcal{E} ^{2} ( C_1 + C_2)\)

 

Illustration Questions

Two capacitors of capacitance \(C_1=3\,\mu F\)  and  \(C_2=4\,\mu F\)  are connected in parallel across the battery of emf \(\mathcal{E}=2\,V\), as shown in figure. Calculate the energy stored in each capacitor.

A \(3\,\mu J,\,4\,\mu J\)

B \(2\,\mu J,\,6\,\mu J\)

C \(6\,\mu J,\,8\,\mu J\)

D \(6\,\mu J,\,3\,\mu J\)

×

Energy stored in capacitor C1

\(U_1=\dfrac{1}{2}C_1 \mathcal{E}^2\)

Or,  \(U_1=\dfrac{1}{2}×3×10^{-6}×(2)^2\)

Or,  \(U_1=6\,\mu J\)

image

Energy stored in capacitor C2

\(U_2=\dfrac{1}{2}C_2 \mathcal{E}^2\)

Or,  \(U_2=\dfrac{1}{2}×4×10^{-6}×(2)^2\)

Or,  \(U_2=8\,\mu J\)

image

Two capacitors of capacitance \(C_1=3\,\mu F\)  and  \(C_2=4\,\mu F\)  are connected in parallel across the battery of emf \(\mathcal{E}=2\,V\), as shown in figure. Calculate the energy stored in each capacitor.

image
A

\(3\,\mu J,\,4\,\mu J\)

.

B

\(2\,\mu J,\,6\,\mu J\)

C

\(6\,\mu J,\,8\,\mu J\)

D

\(6\,\mu J,\,3\,\mu J\)

Option C is Correct

Work Done by Battery

  • Consider a capacitor of capacitance C, having an initial charge Qi .
  • When it is connected with battery of emf \(\mathcal{E}\), the capacitor gets charged up to a final value Qf .

  • Initially till the switch is open, the charge on positive plate is Qi .
  • When switch (S) is closed, the capacitor gets charged up to Qf

            \(Q_f=C\mathcal{E}\)

  • Hence, work done by battery

         \(W=\Delta Q.\mathcal{E}\)

        or, \(W=(Q_f-Q_i)\mathcal{E}\)

        or, \(W=(C\mathcal{E}-Q_i)\mathcal{E}\)

Illustration Questions

A capacitor of capacitance \(C=2\,\mu F\), connected with a battery of e.m.f \(\mathcal{E}=2\,V\), as shown in figure. Calculate the work done by battery when the initial charge present on capacitor before connecting it to battery is \(Q_i=6\,\mu C\).

A \(–2\,\mu J\)

B \(–6\,\mu J\)

C \(–3\,\mu J\)

D \(–4\,\mu J\)

×

The initial charge on capacitor before connecting to battery i.e, switch is open is,

\(Q_i=6\,\mu C\)

image

The maximum charge on capacitor after the battery is connected, i.e, switch is closed.

\(Q_f = C\mathcal{E}\)

or,    \(Q_f = 2 × 10^{-6} × 2 \)

or,    \( Q_f =4\,\mu C\)

 

image

Hence, initial charge is greater than the maximum charge acquired by capacitor after connecting it to battery. 

Qmax < Qi 

Amount of charge supplied by battery,

\(\Delta Q = Q_f - Q_i\)

\(\Delta Q = 4 \,–\, 6 =\, – 2 \mu C\)

 

image

Hence, work done by battery 

\(W = \Delta Q × \mathcal{E}\)

\(or, W = – 2 × 10^{-6} × 2 \)

or,   \(W=\,–4\,\mu J\)

A capacitor of capacitance \(C=2\,\mu F\), connected with a battery of e.m.f \(\mathcal{E}=2\,V\), as shown in figure. Calculate the work done by battery when the initial charge present on capacitor before connecting it to battery is \(Q_i=6\,\mu C\).

image
A

\(–2\,\mu J\)

.

B

\(–6\,\mu J\)

C

\(–3\,\mu J\)

D

\(–4\,\mu J\)

Option D is Correct

Energy Radiated in the Form of Heat

  • Consider a capacitor of capacitance C, connected to a battery with e.m.f. \(\mathcal{E}\), as show in figure. 
  • Work done by battery is 

           \(W = (Q_f - Q_i) × \mathcal{E}\)

 

  •  If initially, capacitor is uncharged,

          \(Q_i = 0\)

         so, Qf = Qmax =  \(C \mathcal{E}\)

  • Hence, work done is 

           W = Qmax × \(\mathcal{E}\) 

or,     W = \(C \mathcal{E} × \mathcal{E}\)

or,     W = \(C\mathcal{E}^{2}\)

  • Also, energy stored in capacitor is 

          \(U = \dfrac{1}{2} C\mathcal{E}^{2}\)

Conclusion :- The amount of energy stored in capacitor is half of work done by battery. It means half of the energy is radiated as heat.

Heat = work done by battery - energy stored in capacitor 

⇒ Heat = \(W -\dfrac{1}{2} C\mathcal{E}^{2}\)

⇒ Heat = \(C\mathcal{E}^{2} - \dfrac{1}{2}C\mathcal{E}^{2}\)

⇒ Heat = \(\dfrac{1}{2}C\mathcal{E}^{2}\)

 

Illustration Questions

An uncharged capacitor having capacitance \(C = 5 \,\mu F \), is connected with a battery of e.m.f. \(\mathcal{E} = 10 \,V\). Calculate the heat produced by battery.

A \(200\, \mu J\)

B \(250\, \mu J\)

C \(300\, \mu J\)

D \(350\, \mu J\)

×

Heat = work done by battery - energy stored in capacitor 

⇒ Heat = \(W- \dfrac{1}{2} C\mathcal{E}^{2}\)

⇒ Heat = \(C\mathcal{E}^{2} - \dfrac{1}{2}C\mathcal{E}^{2}\)

⇒ Heat = \(\dfrac{1}{2}C\mathcal{E}^{2}\)

 

Heat = \(\dfrac {1}{2} × 5 × 10^{-6} × (10)^{2}\)

Heat = \(250× 10^{-6} \)

Heat = \(250\, \mu J\)

An uncharged capacitor having capacitance \(C = 5 \,\mu F \), is connected with a battery of e.m.f. \(\mathcal{E} = 10 \,V\). Calculate the heat produced by battery.

A

\(200\, \mu J\)

.

B

\(250\, \mu J\)

C

\(300\, \mu J\)

D

\(350\, \mu J\)

Option B is Correct

Work Done by Battery when its Polarity is Interchanged

  • Consider a capacitor having capacitance C is connected to a battery with e.m.f. \(\mathcal{E}\) as shown in figure.   

  • Initially, the capacitor is uncharged,  i.e., Qi = 0.
  • When capacitor is connected to battery, the capacitor gets fully charged up to  maximum value (Qmax). 

    Qf = Qmax = \(C\mathcal{E}\)

  • Hence, the charge flow through battery is 

    \(\Delta Q = Q_f – Q_i\)

    or,   \(\Delta Q = Q_{max} – 0\)

    or,   \(\Delta Q = Q_{max} =C \mathcal{E}\)    

 

  • Work done by battery,

W1 = \(\Delta Q × \mathcal{E} = C\mathcal{E} ^{2}\)

  • Now, if the polarity of battery is reversed. 
  • Before reversing the polarity of battery, charge on negative plate is – Q ,           

     Q'i = – Q

 

 

 

  • After reversing the battery charge on negative plate is + Q.

  • Final charge on capacitor, 

Q'= + Q = + \(C \mathcal{E}\)

  • Hence, charge flow through battery is 

    \(\Delta\)Q' = Q'f – Q'i

    or,     \(\Delta\)Q' = + Q – (– Q)

    or,     \(\Delta\)Q' = 2 Q

  • Hence, Work done by battery 

    W2 = \(\Delta\)Q' × \(\mathcal{E}\) \(= 2 \,Q ×\mathcal{E}\)

                            = + \(2 C\mathcal{E}^{2}\)    

Illustration Questions

An uncharged capacitor having capacitance \(C = 3 \,\mu F\), is connected to a battery \(\mathcal{E} = 2 \,V\). Calculate work done by the battery if the polarity of battery is reversed.

A \(6\, \mu J\)

B \(4\, \mu j\)

C \(3\, \mu J\)

D \(24\, \mu J\)

×

Initially, the capacitor is uncharged, 

Qi = 0

image

When capacitor is connected to battery, the capacitor gets fully charge up to maximum value (Qmax). 

Qf = Qmax = \(C\mathcal{E}\)

Qf = 3 × 10-6 × 2 

Qf = \(6\, \mu C\)

image

Hence, the charge flow through battery is 

\(\Delta\)Q = Qf – Qi

or,   \(\Delta\)Q = Qmax – 0

or,   \(\Delta\)Q = Qmax = \(C\mathcal{E}\)    

\(\Delta\)Q = 3 × 10-6 × 2 

\(\Delta\)\(Q = 6 \,\mu C\)

image

Work done by battery,

W1 = \(\Delta Q × \mathcal{E} =C\mathcal{E} ^{2}\)

W1 = 3 × 10-6 × (2)2

W1 = \(12\, \mu J\)

image

Now, the the polarity of battery is reversed.

image

Before reversing the polarity of battery, charge on negative plate is – Q is,            

 Q'i = – \(6 \mu c\)

 

image image

After reversing the battery, charge on negative plate is \(+ 6 \,\mu C\).

image image

Final charge on capacitor, 

Q'= + Q = + \(C\mathcal{E}\)

Q'f = 3 × 10-6 × 2

Q'f = \(6 \,\mu C\)

image

Hence, charge flow through battery is 

\(\Delta\)Q' = Q'f – Q'i

or,     \(\Delta\)Q' = + Q – (– Q)

or,     \(\Delta\)Q' = 2 Q

⇒ \(\Delta\)Q' = \(2 × 6 \mu C\)

⇒ \(\Delta\)Q' \(12 \mu C\)

image

Hence, Work done by battery, 

W2 = \(\Delta\)Q' × \(\mathcal{E}\)= 2 Q × \(\mathcal{E}\)

=  \(2 C\mathcal{E}^{2}\)    

W2 = 2 × 3 × 10-6 × (2)2

W2 = \(24 \,\mu J\) 

image

An uncharged capacitor having capacitance \(C = 3 \,\mu F\), is connected to a battery \(\mathcal{E} = 2 \,V\). Calculate work done by the battery if the polarity of battery is reversed.

image
A

\(6\, \mu J\)

.

B

\(4\, \mu j\)

C

\(3\, \mu J\)

D

\(24\, \mu J\)

Option D is Correct

Potential Energy across a Capacitor 

  • Consider an uncharged capacitor C, connected to a battery with e.m.f. \(\mathcal{E}\), as shown in figure.

  • Potential energy of the charged capacitor is 

          \(U= \dfrac {1}{2} C\mathcal{E}^{2}\)

  • Now, this charged capacitor is connected with another battery of same e.m.f. but polarity is reversed. 

 

  •  The initial potential energy when switch is open 

          \(U_i = \dfrac {1}{2} C\mathcal{E}^{2}\)

  • The final potential energy when switch is closed

         \(U_f = \dfrac {1}{2} C\mathcal{E}^{2}\)

  • Hence, change in potential energy 

\(\Delta\)U = Uf – Ui

or,  \(\Delta U = \dfrac{1}{2} C\mathcal{E}^{2} - \dfrac{1}{2}C\mathcal{E}^{2}\)

or, \(\Delta U = 0 \)

Note :- The energy stored in the capacitor remains same even if the polarity of the battery is reversed. 

Thus, work done by battery appears as heat.

Illustration Questions

An uncharged capacitor of capacitance \(C =4 \,\mu F \) is connected with battery \(\mathcal{E} = 3 \,V\). Calculate the change in potential energy when the polarity of battery is reversed.

A 18 J

B Zero 

C – 18 J

D 36 J

×

Initial potential energy when battery is connected as shown in figure.

\(U_i = \dfrac {1}{2} C\mathcal{E}^{2}\)

⇒  \(U_i =\dfrac{1}{2}× 4 × 10^{-6}× (3)^{2}\)

⇒  \(U_i=18\, \mu J\)

image

Final potential energy when battery is reversed,

\(U_f = \dfrac {1}{2} C\mathcal{E}^{2}\)

⇒  \(U_f =\dfrac{1}{2}× 4 × 10^{-6}× (3)^{2}\) 

⇒ \(U_f = 18\, \mu J\)

 

image

So, change in potential energy is given as 

\(\Delta U = U_f - U_i\)

\(\Delta U\) = \(18 \,\mu J\) – \(18\, \mu J\)

\(\Delta U=0\)

An uncharged capacitor of capacitance \(C =4 \,\mu F \) is connected with battery \(\mathcal{E} = 3 \,V\). Calculate the change in potential energy when the polarity of battery is reversed.

A

18 J

.

B

Zero 

C

– 18 J

D

36 J

Option B is Correct

Work Done by External Force

  • Consider an uncharged capacitor, as shown in figure.  

         Charge on capacitor Q = 0

  • When switch is closed, the capacitor gets fully charged up to Q amount.  

Charge on capacitor Q = \(C\mathcal{E}\)

  • Assume that the external force is applied on plates of capacitor to increase the separation between them from d to 2d. 

  • Then capacitance of capacitor ,\(C^{'} = \dfrac{\mathcal{E} A}{d^{'}}\) 

When d increase to 2d

 \(C^{'} = \dfrac{\mathcal{E} A}{2 d} = \dfrac {C}{2}\)

  • Hence, charge on capacitor will be 

\(Q^{'} = C^{'} V\)

\(Q^{'} = \dfrac {C}{2} V = \dfrac {Q}{2}\)

  • Initial potential energy 

            \(U_i = \dfrac {1}{2} C\mathcal{E} ^{2}\)

  • Final potential energy 

         \(U_f = \dfrac{1}{2} C^{'}\mathcal{E}^{2}\)

       ⇒ \(U_f = \dfrac{1}{2}(\dfrac{C}{2})\mathcal{E} ^{2}\)

       ⇒ \(U_f = \dfrac{1}{4}\mathcal{E} ^{2}\)

  • Work done in increasing the separation from d to 2d,

Wb = \(\Delta Q .\mathcal{E}\)

⇒ W=\(( Q_f - Q_i ) \mathcal{E}\)

⇒ W\((\dfrac {Q}{2} - Q ) \mathcal{E}\)

⇒ W\(\dfrac{-Q}{2}\mathcal{E}\)

⇒ Wb = \(\dfrac {-1}{2}C\mathcal{E} ^{2}\)

  • Hence, work done by external force can be determined by, 

Ui + Wb = Uf + Wext

⇒ \(\dfrac {1}{2} C\mathcal{E} ^{2} - \dfrac {1}{2} C\mathcal{E} ^{2} = \dfrac {1}{4}C\mathcal{E}^{2} \) + Wext

⇒ Wext \(\dfrac {-1}{4} C\mathcal{E} ^{2}\)

Note :- While moving the plates of capacitor slowly, heat is zero. 

          Heat = 0

Illustration Questions

A charged capacitor (C) is connected to a battery with e.m.f. \(\mathcal{E}\). Chose the incorrect option when the separation between capacitor plate is halved. 

A Wext = \(\dfrac {1}{2}C\mathcal{E}^{2}\)

B WB = \(\dfrac {1}{2}C\mathcal{E} ^{2}\)

C \(\Delta U = \dfrac {1}{2}C\mathcal{E}^{2}\)

D WB = \(C\mathcal{E}^{2}\)

×

When the separation between plates is halved

d' = \(d/2\)

\(C^{'}= \dfrac {\varepsilon _0A}{d^{'}} = \dfrac {{\varepsilon _0A}}{(d/2)}\)

\(C^{'} = \dfrac {2 \varepsilon _0A}{d} = 2C\)

\(C^{'} = 2C\)

Hence, capacitance is doubled.

 

Charge on capacitor plates (Q')

Q' = C' V

Q' = 2 CV

Q' = 2Q

Initial potential energy (Ui)

\(U_i = \dfrac {1}{2} C\mathcal{E} ^{2}\)

Final potential energy (Uf)

\(U_f = \dfrac {1}{2}C^{'} \mathcal{E}^{2}\)

\(U_f = \dfrac {1}{2} (2C) \mathcal{E} ^{2} = C \mathcal{E} ^{2}\)

Work done ( W)

Wb = \(\Delta Q .\mathcal{E}\)

⇒ W=\(( Q_f - Q_i )\mathcal{E}\)

⇒ W\((2 Q - Q) \mathcal{E}\)

⇒ W\(Q \mathcal{E}\)

⇒ W\(C\mathcal{E}^{2}\)

 

 

Work done by external force (Wext)

Ui +Wb = Uf +Wext

\(\dfrac{1}{2} C\mathcal{E}^{2} + C\mathcal{E}^{2} = C\mathcal{E}^{2} \) + Wext

Wext \(\dfrac{1}{2} C\mathcal{E}^{2}\) 

Change in potential energy \((\Delta \;U )\)

\(\Delta U = U_f - U_i\)

\(= C\mathcal{E} ^{2} - \dfrac{1}{2} C\mathcal{E} ^{2}\)

\(\dfrac{1}{2} C\mathcal{E}^{2}\)

Hence, option ( B ) is incorrect.

A charged capacitor (C) is connected to a battery with e.m.f. \(\mathcal{E}\). Chose the incorrect option when the separation between capacitor plate is halved. 

A

Wext = \(\dfrac {1}{2}C\mathcal{E}^{2}\)

.

B

WB = \(\dfrac {1}{2}C\mathcal{E} ^{2}\)

C

\(\Delta U = \dfrac {1}{2}C\mathcal{E}^{2}\)

D

WB = \(C\mathcal{E}^{2}\)

Option B is Correct

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