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
Step-by-step 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.