Parallel Plate Capacitor Dielectric Insertion – Rankers Physics

Parallel Plate Capacitor: Practice Problem & Solution

Assertion (A): A parallel plate capacitor is charged to a potential difference of \( 100\text{V} \), and disconnected from the voltage source. A slab of dielectric is then slowly inserted between the plates. Compared to the energy before the slab was inserted, the energy stored in the capacitor with the dielectric is decreased. Reason (R): When we insert a dielectric between the plates of a capacitor, the induced charges tend to draw in the dielectric into the field (just as neutral objects are attracted by charged objects due to induction). We resist this force while slowly inserting the dielectric, and thus do negative work on the system, removing electrostatic energy from the system.  
Both (A) & (R) are true and the (R) is the correct explanation of the (A)
Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
(A) is true but (R) is false
Both (A) and (R) are false

Short Solution (Best for NEET):

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:

When a dielectric is inserted into a disconnected charged capacitor, the charge \( Q \) remains constant. The capacitance \( C \) increases to \( kappa C_0 \), where \( \kappa \) is the dielectric constant. The energy stored is \( U = \frac{Q^2}{2C} \). Since \( C \) increases, \( U \) decreases. The external agent does negative work, as the dielectric is pulled in by electrostatic forces. This decrease in energy is explained by the work done by the field.

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