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.
\(E_+ = \dfrac{\sigma}{2\epsilon _o} = \dfrac{Q}{2A\epsilon_ o}\)
\(F=-QE\)
\(F= -Q.\dfrac{Q}{2A\epsilon _o}\)
\(F = \dfrac{–Q^{2}}{2A\epsilon_o}\)
\(F = \dfrac{Q^{2}}{2A\epsilon_o}\) ( Force of attraction )
\(W = Fd\)
\(W = \dfrac{Q^{2}d}{2A\epsilon_0}\)
or \(W = \dfrac{Q^{2}}{2 C}\)
\(U = \dfrac{Q^{2}}{2 C}\)
\(\Delta V = \dfrac{q}{C}\)
\(dW=\dfrac{q} {C}dq \)
\(\int\limits ^W_0dW = \int\limits^Q_0 \dfrac {q}{C} dq \)
or \(W = \dfrac{Q^{2}}{2 C}\)
\(U = \dfrac{Q^{2}}{2 C}\)
A Charge is doubled
B Charge is halved
C Potential difference is doubled
D Potential difference is halved
\(U_1 =\dfrac {Q^{2}}{2C_1}\)
\(U_2 =\dfrac {Q^{2}}{2C_2}\)
U = U_{1} + U_{2}
\(\Rightarrow U = \dfrac {Q^{2}}{2C_1} + \dfrac {Q^{2}}{2 C_2}\)
\(\Rightarrow U = Q^{2} ( \dfrac {1}{2C_1} + \dfrac{1}{2C_2})\)
\(\Rightarrow U = \dfrac {Q^{2}}{2} \left (\dfrac {1}{C_1} + \dfrac{1}{C_2} \right)\)
A \(8\,\mu J,\,12\,\mu J\)
B \(6\,\mu J,\,12\,\mu J\)
C \(4\,\mu J,\,12\,\,\mu J\)
D \(12\,\mu J,\,3\,\mu J\)
\(U_1 = \dfrac {1}{2} C_1\mathcal{E} ^{2}\)
\(U_2 = \dfrac {1}{2} C_2 \mathcal{E} ^{2}\)
\(U = U_1 + U_2\)
\(\Rightarrow U = \dfrac{1}{2} C_1 \mathcal{E}^{2} + \dfrac {1}{2} C_2 \mathcal{E}^{2} \)
\(\Rightarrow U = \dfrac{1}{2}\mathcal{E} ^{2} ( C_1 + C_2)\)
A \(3\,\mu J,\,4\,\mu J\)
B \(2\,\mu J,\,6\,\mu J\)
C \(6\,\mu J,\,8\,\mu J\)
D \(6\,\mu J,\,3\,\mu J\)
\(Q_f=C\mathcal{E}\)
\(W=\Delta Q.\mathcal{E}\)
or, \(W=(Q_f-Q_i)\mathcal{E}\)
or, \(W=(C\mathcal{E}-Q_i)\mathcal{E}\)
A \(–2\,\mu J\)
B \(–6\,\mu J\)
C \(–3\,\mu J\)
D \(–4\,\mu J\)
\(W = (Q_f - Q_i) × \mathcal{E}\)
\(Q_i = 0\)
so, Q_{f} = Q_{max} = \(C \mathcal{E}\)
W = Q_{max} × \(\mathcal{E}\)
or, W = \(C \mathcal{E} × \mathcal{E}\)
or, W = \(C\mathcal{E}^{2}\)
\(U = \dfrac{1}{2} C\mathcal{E}^{2}\)
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 = \(\dfrac{1}{2}C\mathcal{E}^{2}\)
A \(200\, \mu J\)
B \(250\, \mu J\)
C \(300\, \mu J\)
D \(350\, \mu J\)
When capacitor is connected to battery, the capacitor gets fully charged up to maximum value (Q_{max}).
Q_{f} = Q_{max} = \(C\mathcal{E}\)
Hence, the charge flow through battery is
\(\Delta Q = Q_f – Q_i\)
or, \(\Delta Q = Q_{max} – 0\)
or, \(\Delta Q = Q_{max} =C \mathcal{E}\)
W_{1} = \(\Delta Q × \mathcal{E} = C\mathcal{E} ^{2}\)
Before reversing the polarity of battery, charge on negative plate is – Q ,
Q'_{i} = – Q
Q'_{f }= + 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
W_{2} = \(\Delta\)Q' × \(\mathcal{E}\) \(= 2 \,Q ×\mathcal{E}\)
= + \(2 C\mathcal{E}^{2}\)
A \(6\, \mu J\)
B \(4\, \mu j\)
C \(3\, \mu J\)
D \(24\, \mu J\)
\(U= \dfrac {1}{2} C\mathcal{E}^{2}\)
\(U_i = \dfrac {1}{2} C\mathcal{E}^{2}\)
\(U_f = \dfrac {1}{2} C\mathcal{E}^{2}\)
\(\Delta\)U = U_{f} – U_{i}
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.
Charge on capacitor Q = 0
Charge on capacitor Q = \(C\mathcal{E}\)
When d increase to 2d
\(C^{'} = \dfrac{\mathcal{E} A}{2 d} = \dfrac {C}{2}\)
\(Q^{'} = C^{'} V\)
\(Q^{'} = \dfrac {C}{2} V = \dfrac {Q}{2}\)
\(U_i = \dfrac {1}{2} C\mathcal{E} ^{2}\)
\(U_f = \dfrac{1}{2} C^{'}\mathcal{E}^{2}\)
⇒ \(U_f = \dfrac{1}{2}(\dfrac{C}{2})\mathcal{E} ^{2}\)
⇒ \(U_f = \dfrac{1}{4}\mathcal{E} ^{2}\)
W_{b} = \(\Delta Q .\mathcal{E}\)
⇒ W_{b }=\(( Q_f - Q_i ) \mathcal{E}\)
⇒ W_{b }= \((\dfrac {Q}{2} - Q ) \mathcal{E}\)
⇒ W_{b }= \(\dfrac{-Q}{2}\mathcal{E}\)
⇒ W_{b} = \(\dfrac {-1}{2}C\mathcal{E} ^{2}\)
U_{i} + W_{b} = U_{f} + W_{ext}
⇒ \(\dfrac {1}{2} C\mathcal{E} ^{2} - \dfrac {1}{2} C\mathcal{E} ^{2} = \dfrac {1}{4}C\mathcal{E}^{2} \) + W_{ext}
⇒ W_{ext }= \(\dfrac {-1}{4} C\mathcal{E} ^{2}\)
Note :- While moving the plates of capacitor slowly, heat is zero.
Heat = 0
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 = \(C\mathcal{E}^{2}\)