Question 573: Electrostatics & Capacitance

Question

When the battery across the plates of a charged condenser is disconnected and a dielectric slab is introduced between its plates then the energy stored

Accepted Answer

To understand what happens to the energy stored in a charged capacitor when a dielectric slab is introduced between its plates (after disconnecting the battery), let's break down the concepts involved.
A capacitor stores energy in the electric field between its plates. The energy stored in a capacitor is given by the formula:
$$ U = \frac{1}{2} \frac{Q^2}{C} $$
where:
U
is the energy stored.
Q
is the charge on the capacitor.
C
is the capacitance of the capacitor.
The capacitance of a capacitor is given by:
$$ C = \frac{\varepsilon_0 A}{d} $$
where:
$\varepsilon_0$
is the permittivity of free space.
A
is the area of the plates.
d
is the distance between the plates.
When a dielectric slab with dielectric constant
ÃÂº
is introduced between the plates, the new capacitance
C'
becomes:
$$ C' = \kappa C $$
Since the battery is disconnected, the charge
Q
on the capacitor remains constant. Let's understand the impact of the increased capacitance (due to the dielectric slab) on the stored energy.
The new energy stored
U'
can be expressed as:
$$ U' = \frac{1}{2} \frac{Q^2}{C'} $$
Substituting $C' = \kappa C$, we get:
$$ U' = \frac{1}{2} \frac{Q^2}{\kappa C} $$
This equation shows that:
$$ U' = \frac{U}{\kappa} $$
We know that $\kappa > 1$, therefore:
$$ U' < U $$
This indicates that the energy stored in the capacitor decreases when a dielectric slab is introduced between its plates (after disconnecting the battery).
So, the correct answer is:
Option D: decreases

Answer

becomes infinity

Answer

does not change

Answer

increases

Answer

decreases

Electrostatics & Capacitance

When the battery across the plates of a charged condenser is disconnected and a dielectric slab is introduced between its plates then the energy stored

Choose the correct answer: