Q138 / 160 · TS EAMCET 2003 · Physics

A parallel plate capacitor of capacity $C_0$ is charged to a potential $V_0$. (i) The energy stored in the capacitor when the battery is disconnected and the plate separation is doubled is $E_1$. (ii) The energy stored in the capacitor when the charging battery is kept connected and the separation between the capacitor plates is doubled is $E_2$. Then, $E_1 / E_2$ value is:

  • Correct answer
    A
    4
  • B
    $3 / 2$
  • C
    2
  • D
    $1 / 2$

Solution

Capacitance of parallel plate capacitor, $C_0=\frac{\varepsilon_0 A}{d}$ where, $A=$ area of the plates, $d=$ separation between the plates Charge stored in the capacitor, $Q=C_0 V_0$ When battery is disconnected, then charge remains same. So, energy, $E_1=\frac{1}{2} \frac{Q^2}{C}$ $C=$ capacitance when plate separation is doubled So, $\quad C_1=\frac{C_0}{2}$ $E_1=\frac{1}{2} \frac{Q^2}{C_0 / 2}=\frac{Q^2}{C_0}=\frac{C_0^2 V_0^2}{C_0}=C_0 V_0^2$ When battery is connected, then Energy, $E_2=\frac{1}{2} C V_0^2$ where, $E_2=\frac{1}{2} \frac{C_0}{2} V_0^2=\frac{1}{4}\left(C_0 V_0^2\right)$ $\therefore \quad \frac{E_1}{E_2}=\frac{C_0 V_0^2}{\frac{1}{4} C_0 V_0^2}=\frac{1}{4}$ $E_1: E_2=4: 1$

Hence, the correct answer is option: 'A'.

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