Parallel Plate Capacitor: Practice Problem & Solution

Short Solution (Best for NEET):

To solve this problem, we apply the core principles of Parallel Plate Capacitor. Understanding the underlying formula is key to arriving at the correct answer below:

To solve this, let's calculate the percentage of energy dissipated when the switch

$S$

is turned to position 2.

Given:

• 
  $C_1 = 2 \, \mu\text{F}$ 
• 
  $C_2 = 8 \, \mu\text{F}$ 
• Initial voltage across
  $C_1 = V$ 

Key Concept:

When two capacitors are connected, the redistribution of charge causes energy loss. The energy dissipated can be calculated using the direct formula:

$$\text{Energy dissipated} = \frac{1}{2} \frac{C_1 C_2}{C_1 + C_2} (V_1 - V_2)^2$$

Here:

• 
  $V_1$ 

  is the initial voltage of $C_1$ 

  .

• 
  $V_2 = 0$ 

  (initially, $C_2$ 

  is uncharged).

Step-by-step Calculation:

1. Common Voltage after Connection: The final voltage across both capacitors is: 

   $$V_{\text{common}} = \frac{C_1 V}{C_1 + C_2} = \frac{2V}{2 + 8} = \frac{V}{5}.$$ 

2. Initial Energy Stored in
   $C_1$ 

   : 

   $$E_{\text{initial}} = \frac{1}{2} C_1 V^2 = \frac{1}{2} (2 \times 10^{-6}) V^2 = 10^{-6} V^2 \, \text{J}.$$ 

3. Final Energy Stored: The total energy stored after redistribution is: 

   $$E_{\text{final}} = \frac{1}{2} (C_1 + C_2) V_{\text{common}}^2 = \frac{1}{2} (2 + 8) \left(\frac{V}{5}\right)^2.$$Substituting values:

    

   $$E_{\text{final}} = \frac{1}{2} (10) \left(\frac{V^2}{25}\right) = \frac{10 V^2}{50} = \frac{V^2}{5} \times 10^{-6} \, \text{J}.$$ 

4. Energy Dissipated: 

   $$E_{\text{dissipated}} = E_{\text{initial}} - E_{\text{final}} = 10^{-6} V^2 - \frac{10^{-6} V^2}{5} = \frac{4}{5} \times 10^{-6} V^2.$$ 

5. Percentage of Energy Dissipated: 

   $$\% \text{Dissipated} = \frac{E_{\text{dissipated}}}{E_{\text{initial}}} \times 100 = \frac{\frac{4}{5} \times 10^{-6} V^2}{10^{-6} V^2} \times 100 = 80\%.$$ 

Final Answer:

The percentage of energy dissipated is 80%.