Question

A capacitor of capacitance \(2\,\mu\text{F}\) is charged to \(10\,\text{V}\). Energy stored is:

• A. $50\mu\text{J}$
• B. $100\mu\text{J}$
• C. $200\mu\text{J}$
• D. $400\mu\text{J}$

Hint:

To perform calculations quickly, you can keep the capacitance in microfarads ($\mu\text{F}$). The resulting energy will automatically be in microjoules ($\mu\text{J}$):
\[ U = \frac{1}{2} \cdot 2\mu\text{F} \cdot 10^2 = 100\mu\text{J} \]

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
The question asks to calculate the electrostatic potential energy ($U$) stored in a parallel-plate capacitor of a given capacitance when it is charged to a specific potential difference.

Step 2: Key Formula or Approach:
The potential energy stored in a charged capacitor can be calculated using the formula:
\[ U = \frac{1}{2} C V^2 \] where $C$ is the capacitance and $V$ is the potential difference applied across the plates.

Step 3: Detailed Explanation:

• Let us identify the given values:
Capacitance, $C = 2\mu\text{F} = 2 \times 10^{-6}\text{ F}$
Potential difference, $V = 10\text{ V}$

• Substituting these values directly into the energy formula:
\[ U = \frac{1}{2} \cdot (2 \times 10^{-6}) \cdot (10)^2 \]
• Simplifying the calculations:
\[ U = 10^{-6} \cdot 100 \] \[ U = 100 \times 10^{-6}\text{ J} \]
• Since $10^{-6}\text{ J}$ is equal to $1\mu\text{J}$, we can write the stored energy as:
\[ U = 100\mu\text{J} \]
• This potential energy is stored in the electric field created between the plates of the capacitor.

Step 4: Final Answer:
The electrostatic potential energy stored in the capacitor is $100\mu\text{J}$.