Initial capacitance with vacuum:
\[ C = \frac{\epsilon_0 A}{d} \]
Thickness of dielectric slab = \( \frac{d}{3} \)
Remaining vacuum thickness = \( \frac{2d}{3} \)
This configuration behaves as two capacitors connected in series.
\[ C_1 = \frac{K\epsilon_0 A}{d/3} = \frac{3K\epsilon_0 A}{d} \]
\[ C_2 = \frac{\epsilon_0 A}{2d/3} = \frac{3\epsilon_0 A}{2d} \]
\[ \frac{1}{C'} = \frac{1}{C_1} + \frac{1}{C_2} \]
\[ \frac{1}{C'} = \frac{d}{3K\epsilon_0 A} + \frac{2d}{3\epsilon_0 A} \]
\[ C' = \frac{3KC}{2K+1} \]
Therefore correct answer is (D).
Capacitance is one of the most fundamental concepts in electrostatics and plays a major role in JEE Main and JEE Advanced examinations. A capacitor is a device that stores electric charge and electrical energy in an electric field. For a parallel plate capacitor with vacuum between plates, capacitance depends only on geometry and permittivity of free space.
\[ C = \frac{\epsilon_0 A}{d} \]
This formula shows capacitance increases with plate area and decreases with separation.
When a dielectric material is introduced between the plates, the electric field inside the capacitor changes due to polarization of molecules. The dielectric reduces effective electric field by factor K. Relative permittivity K is defined as:
\[ K = \frac{\epsilon}{\epsilon_0} \]
If the dielectric completely fills the space, new capacitance becomes KC.
However, when dielectric fills only part of the gap along thickness direction, the system must be treated as two capacitors in series. This is because electric field lines pass sequentially through vacuum and dielectric regions.
In such cases, equivalent capacitance is found using series combination:
\[ \frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} \]
This principle is extremely important in competitive exams.
Physical understanding: Inside dielectric, electric field reduces to E/K. But displacement field remains continuous. Boundary conditions between dielectric and vacuum ensure normal component of displacement is same.
Energy stored in capacitor:
\[ U = \frac{1}{2}CV^2 \]
If capacitor remains connected to battery, voltage remains constant and inserting dielectric increases capacitance, thus increasing stored energy. If capacitor is isolated, charge remains constant and energy decreases.
This distinction is very important for JEE Advanced problems.
Capacitors are used in filtering circuits, oscillators, timing devices, power supplies and electronic communication systems. Understanding dielectric insertion builds foundation for electric polarization, Gauss’s law in dielectrics and Maxwell’s equations.
Common mistakes students make:
• Using KC directly without considering thickness fraction
• Confusing series and parallel combination cases
• Ignoring physical reasoning and memorizing formula
• Algebraic mistakes in simplification
Exam relevance:
Electrostatics contributes 3–4 questions in JEE Main. Capacitor modification questions are very common because they test conceptual clarity and algebraic accuracy.
Advanced perspective: When dielectric is inserted, polarization charges appear on surfaces. These bound charges reduce net electric field. In microscopic view, molecular dipoles align in field direction. This is polarization mechanism.
Understanding this topic deeply helps in mastering electric displacement vector D, polarization vector P and boundary conditions.
Students should practice variations involving multiple dielectric slabs, insertion by area fraction and removal of dielectric.
With strong conceptual clarity, such questions become scoring and less time-consuming in exam.
This educational content is prepared for practice and learning purposes for competitive exams such as JEE Main, JEE Advanced and NEET. All problems are solved using standard physics principles and formulas. Students are advised to verify final answers during exam preparation.