Step 1: Understanding the Concept:
A capacitor is an electronic component that stores electrical energy in an electric field between its plates.
As charge builds up on the plates of a capacitor, work must be done by the external source (like a battery) to move more charge onto the plates against the existing electric field.
This work done is stored as electrostatic potential energy.
The amount of energy stored depends on the capacity of the capacitor to store charge (capacitance, \(C\)) and the potential difference (voltage, \(V\)) across it.
Step 2: Key Formula or Approach:
The standard formula for the energy (\(U\)) stored in a capacitor is:
\[ U = \frac{1}{2} C V^{2} \]
Alternatively, if the charge (\(Q\)) is known, it can be written as \( U = \frac{Q^{2}}{2C} \) or \( U = \frac{1}{2}QV \).
In this problem, since \( C \) and \( V \) are given, the first formula is the most direct approach.
Step 3: Detailed Explanation:
Let us identify the given values and perform the conversion to SI units:
Capacitance, \( C = 2 \mu\text{F} \). We must convert microfarads (\( \mu\text{F} \)) to Farads (F).
\( 1 \mu\text{F} = 10^{-6} \text{ F} \), so \( C = 2 \times 10^{-6} \text{ F} \).
Potential difference, \( V = 10 \text{ V} \).
Now, substitute these into the energy equation:
\[ U = \frac{1}{2} \times (2 \times 10^{-6}) \times (10)^{2} \]
Simplify the expression step-by-step:
The \( \frac{1}{2} \) and the factor of \( 2 \) cancel each other out:
\[ U = (1 \times 10^{-6}) \times 100 \]
\[ U = 100 \times 10^{-6} \text{ Joules} \]
In scientific terms, \( 10^{-6} \text{ Joules} \) is equal to one microjoule (\( \mu\text{J} \)).
Therefore:
\[ U = 100 \mu\text{J} \]
This energy represents the total potential energy that can be recovered if the capacitor is discharged through an external circuit.
Step 4: Final Answer:
The electrostatic energy stored in the capacitor is \( 100 \mu\text{J} \).