A 13.0 μF capacitor is connected to a power supply that keeps a constant potential difference of 24.0 VV across the plates. A piece of material having a dielectric constant of 3.75 is placed between the plates, completely filling the space between them.

(a) How much energy is stored in the capacitor before and after the dielectric is inserted?
(b) By how much did the energy change during the insertion? Did it increase or decrease?

Energy stored in the capacitor before the dielectric was inserted=

3.744×10-3 J

Energy stored in the capacitor after the dielectric was inserted=

14.0×10-3 J

Change in energy due to dielectric

10.3×10-3 J or 10mJ

The energy stored in the capacitor increased due to the insertion of the dielectric material.

Explanation:

The energy stored in the capacitor is calculated from the formula

U= 1/2CV^2

Where:

U=energy stored in the capacitor

C=capacitance of the capacitor

V= voltage.

The energy stored in the capacitor before the insertion of the dielectric is shown as U while the energy stored in the capacitor after the insertion of the dielectric is designated U'.

Since U'>U from the solution attached, the change in energy due to insertion of the dielectric ∆U= U'-U

Hence the answer shown. See attached image for detailed solution.

Given Information:

Capacitance = C = 13  μF

Potential difference = V = 24 V

Dielectric constant = de = 3.75

Required Information:

(a) Energy stored before and after adding dielectric = ?

(b) change in energy = ?

(a) E₁ = 3.74x10⁻³ F and E₂ = 14.04x10⁻³ F

(b) ΔE = 10.3x10⁻³ F increased

Explanation:

The energy stored in a capacitor is given by

E = ½CV²

Where C is the capacitance and V is the potential difference

Energy before the dielectric:

E₁ = ½CV²

E₁ = ½(13x10⁻⁶)(24)²

E₁ = 3.74x10⁻³ F

Energy after the dielectric:

E₂ = ½CdeV²

E₂ = ½(13x10⁻⁶)(3.75)(24)²

E₂ = 14.04x10⁻³ F

Change in energy:

ΔE = E₂ - E₁

ΔE = 14.04x10⁻³ - 3.74x10⁻³

ΔE = 10.3x10⁻³ F

The energy increased after adding the dielectric material.

b

explanation: