Answer:
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 = ?
Answer:
(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.
