Question

Five identical large conducting plates each of area A are placed parallel to each other at separation d. Plates 1 and 3 are connected by thin conducting wire and plate 2 and 5 are connected by another thin conducting wire. The junction of plate 1 and 3 and plate 4 are connected by a battery of emf V as shown.

The system is in steady state. Now, plate 4 is moved upward slowly so that it comes in contact with plate 5. The amount of work done by battery during motion of plate 4 is x mJ.

(Here $\frac{\epsilon_o A}{d} = 30 \mu F, V=10 \text{ volts}$). Find the value of x.

Answer 1

2 mJ

Answer 2

Solution Outline:

1. Identify Equipotential Groups:

   • Before moving plate 4:
     Plates 1 and 3 are connected (potential = V).
     Plates 2 and 5 are connected (potential = φ, unknown).
     Plate 4 is at 0 (battery negative) since the battery is connected between junction (plates 1 & 3) and plate 4.

2. Capacitors and Voltages:

   The four gaps (each of capacitance

   $$C_0 = \frac{\epsilon_0A}{d} = 30\,\mu F$$

   ) have voltages:

   • Between 1 and 2: $V - \phi$
   • Between 2 and 3: $\phi - V$ (magnitude $V-\phi$)
   • Between 3 and 4: $V - 0 = V$
   • Between 4 and 5: $0 - \phi = -\phi$ (magnitude $|\phi|$)

3. Conservation of Charge on the Isolated Group (Plates 2 and 5):

   • On capacitor 1–2, plate 2 gets charge: $-C_0(V-\phi)$.
   • On capacitor 2–3, plate 2 gets charge: $+C_0(\phi-V) = -C_0(V-\phi)$.

   Thus, plate 2 charge $Q_2 = -2C_0(V-\phi)$.

   • On capacitor 4–5, plate 5 gets $+C_0\phi$.

   Since plates 2 and 5 are connected and initially uncharged:

   $$-2C_0(V-\phi) + C_0\phi = 0 \quad\Longrightarrow\quad -2V+3\phi=0 \quad\Longrightarrow\quad \phi=\frac{2}{3}V$$

4. Calculate Energy Stored Initially:

   • Between 1 and 2: $$U_{12}=\frac{1}{2}C_0(V-\phi)^2=\frac{1}{2}C_0\left(\frac{V}{3}\right)^2=\frac{C_0V^2}{18}$$
   • Between 2 and 3: Same as $U_{12}$ $\Rightarrow \frac{C_0V^2}{18}$.
   • Between 3 and 4: $$U_{34}=\frac{1}{2}C_0V^2$$
   • Between 4 and 5: $$U_{45}=\frac{1}{2}C_0\left(\frac{2V}{3}\right)^2=\frac{2C_0V^2}{9}$$

   Total initial energy:

   $$U_i=\frac{C_0V^2}{18}+\frac{C_0V^2}{18}+\frac{1}{2}C_0V^2+\frac{2C_0V^2}{9}=\frac{5}{6}C_0V^2$$

5. After Plate 4 Contacts Plate 5:

   The connection now makes plates 2, 4, and 5 equipotential. With the battery still connected between plates 1 & 3 (at V) and the group (2,4,5), the battery forces:

   $$V_{\text{group}} = 0.$$

   The remaining capacitors are:

   • Between plate 1 and 2: $U=\frac{1}{2}C_0V^2$.
   • Between plate 2 and 3: $U=\frac{1}{2}C_0V^2$.
   • Between plate 3 and (4,5): $U=\frac{1}{2}C_0V^2$.

   Thus, total final energy:

   $$U_f=\frac{3}{2}C_0V^2.$$

6. Work Done by Battery:

   The work done by the battery equals the increase in stored energy:

   $$W=U_f-U_i=\left(\frac{3}{2}-\frac{5}{6}\right)C_0V^2=\frac{2}{3}C_0V^2.$$

   Substituting:

   $$C_0 = 30\times10^{-6}\,F,\quad V = 10\,V,$$ $$W = \frac{2}{3}\times 30\times10^{-6}\times 100 = 0.002\,J = 2\, \text{mJ}.$$

Summary:

• Find the unknown potential $\phi$ using charge conservation for plates 2 & 5. Calculate stored energies in all capacitors initially and after plate 4 contacts plate 5. The battery work equals the difference, which comes out to $\frac{2}{3}C_0V^2 = 2 \text{ mJ}$.