Parallel Plate Capacitor: Practice Problem & Solution

Solution Explained:

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:

Let's analyze the problem step by step and derive why the potential difference is 10 V.

1. Capacitors in Parallel (Initial Condition)

When the 10 capacitors, each with capacitance

$C$

, are connected in parallel:

• The equivalent capacitance is:
  $$C_{\text{parallel}} = 10C$$ 
• The total charge stored when connected to a battery of potential
  $V$ 

  is: $$Q = C_{\text{parallel}} \cdot V = (10C) \cdot V = 10CV$$ 

2. Capacitors in Series (Reconfigured System)

After disconnecting the battery, the capacitors are joined in series:

• The equivalent capacitance for 10 capacitors in series is:
  $$C_{\text{series}} = \frac{C}{10}$$ 
• The charge
  $Q$ 

  stored on the series combination remains the same (as charge is conserved): $$Q_{\text{series}} = Q = 10CV$$ 

3. Potential Across the Series Combination

The potential difference across a capacitor or combination of capacitors is related to the charge

$Q$

and capacitance

$C$

:

$$V_{\text{series}} = \frac{Q}{C_{\text{series}}}$$

Substitute

$Q = 10CV$

and

$C_{\text{series}} = \frac{C}{10}$

:

$$V_{\text{series}} = \frac{10CV}{\frac{C}{10}} = V$$

Thus, the potential across the series combination is

$V = 10V$

,.

Final Answer:

The potential of the series combination is:

$$\boxed{10V}$$