A, B, C, D, E, F are conducting plates each of area A and any two consecutive plates separated by a distance d. The net energy stored in the system after the switch S is closed is?

Steps to Solve

1. Determine the Total Capacitance of the System

The plates A, B, C, D, E, F form a series and parallel network of capacitors. There are three capacitors in series (A and B, C and D, E and F) and two additional capacitors in parallel (the total capacitance from the series is combined with that of the parallel).

For two capacitors in series, the total capacitance $C_s$ can be calculated as: $$ \frac{1}{C_s} = \frac{1}{C_1} + \frac{1}{C_2} $$

For the two capacitors, the capacitance can be expressed as: $$ C = \frac{\varepsilon_0 A}{d} $$

Using this for A and B, for example: $$ C_{A,B} = \frac{\varepsilon_0 A}{d} $$ Thus, for the series combination it’s: $$ \frac{1}{C_s} = \frac{1}{C} + \frac{1}{C} = \frac{2}{C} \implies C_s = \frac{C}{2} $$

2. Calculate the Equivalent Capacitance

Next, total capacitance of the entire setup including all plate combos: $$ C_{total} = C_s + C_s = \frac{C}{2} + \frac{C}{2} = C $$

As this pattern continues, we calculate the individual contributions until arriving at: $$ C_{total} = 3 \times \frac{C}{2} = \frac{3}{2}C $$

3. Calculate the Energy Stored

The energy stored in a capacitor can be calculated using the formula: $$ U = \frac{1}{2} C_{total} V^2 $$

Substituting the values: $$ U = \frac{1}{2} \cdot \frac{3 \varepsilon_0 A}{2d} \cdot (180)^2 $$

Calculating the final value gives you the total energy stored: $$ U = \frac{3 \varepsilon_0 A}{4d} \cdot 32400 $$

4. Final Result

The final energy ensures energy conservation principles applied correctly after the switch is closed.