##
**Energy Stored In a Capacitor In The Presence Of Dielectric**

Energy Stored In a Capacitor In The Presence Of Dielectric |

The energy stored in a capacitor with a dielectric is given as

**U**=1/2 (CV)² →

**1**

Here

**C**= Capacitance = ∈ A/d

And

**V**= E.d

Putting the above values in Equation.1

**U**= 1/2 (∈A/d ) (E.d)²

**U**= 1/2 ∈ A/d E². d²

**U**= 1/2 ∈ A/d E² (A. d) ⟶

**2**

In the above equation, A.d is the volume between the plates of the capacitor containing the electric field and

**∈E²**= k∈∘E²

**∈**= k∈∘

So,

**U/A.d**= 1/2× k×∈∘×E²

In the above equation,

**U/A.d**is the energy per unit volume.

**U/A.d**= Uₑ

So,

**Uₑ**= 1/2×∈∘×E²

The above equation gives the energy stored per unit volume in a capacitor with a dielectric. It clearly shows that energy stored per unit volume in a capacitor with a dielectric is proportional to the square of the electric field.

##
**Numerical****:**

Two parallel plate capacitors having capacitances **Numerical**

**C₁**=

**C₂**=

**2μF**are connected in a parallel across a 12 volts voltage source (battery).

**To Find**

**(a)**. The charge on each capacitor

**(b)**. The total energy stored in the capacitor.

**Numerical Answer:**

**(a).**

As

**Q**= CV

**Q**= (2μF)(12 volts)

**Q**= 24μC

Since capacitors are connected in parallel to the charge on each capacitor will be the same.

**(b).**

Energy stored in each of the capacitors can be calculated using the relation

**U**=1/2×C×V²

**U**=1/2×(2μF)×(12V)²

**U**=144μJ

Since there are two capacitors so the total energy on the combination can be written as

**Uₜₒₜₐₗ**= 2u

= 2 (144μJ)

**Uₜₒₜₐₗ**=288μJ

**Editors Recommendations:**

Energy Stored In a Capacitor In The Presence Of Dielectric
Reviewed by AB Ultimate Guides
on
July 12, 2020
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