MP Board · Class 12 · Physics · Electrostatic Potential and CapacitanceA capacitor is connected to a battery. A dielectric slab of dielectric constant $K$ is inserted between its plates while the battery remains connected. The energy stored in the capacitor:
Step-by-Step Solution
Since battery remains connected, voltage $V$ is constant. $C' = K C_0$. Initial energy $U_0 = \frac{1}{2} C_0 V^2$ and final energy $U' = \frac{1}{2} C' V^2 = K U_0$. Therefore, energy stored increases $K$ times.
Detailed Options Breakdown
Option : Decreases $K$ times
Incorrect choice. This distractor represents a common misunderstanding of the core principles of Electrostatic Potential and Capacitance.
Option 1: Increases $K$ times (Correct Answer)
Correct choice. Refer to the step-by-step verified solution guidelines above for details.
Option 2: Remains constant
Incorrect choice. This distractor represents a common misunderstanding of the core principles of Electrostatic Potential and Capacitance.
Option 3: Becomes zero
Incorrect choice. This distractor represents a common misunderstanding of the core principles of Electrostatic Potential and Capacitance.
💡 Study Guide: This question tests core syllabus concepts from Electrostatic Potential and Capacitance. For formulas, key summaries, and mock exam reference guides, read the full Electrostatic Potential and Capacitance Revision Notes.