Answer:
0.004056 J
0.0154128 J
Increase
Explanation:
C = Capacitance = 12 μF
V = Potential difference = 26 V
K = Dielectric constant = 3.8
New capacitance is
[tex]C'=KC\\\Rightarrow C'=3.8\times 12\times 10^{-6}\\\Rightarrow C'=0.0000456\ F[/tex]
Energy in the capacitor
[tex]U=\dfrac{1}{2}CV^2\\\Rightarrow U=\dfrac{1}{2}\times 12\times 10^{-6}\times 26^2\\\Rightarrow U=0.004056\ J[/tex]
Before insertion energy 0.004056 J
After insertion
[tex]U'=\dfrac{1}{2}C'V^2\\\Rightarrow U'=\dfrac{1}{2}\times 0.0000456\times 26^2\\\Rightarrow U'=0.0154128\ J[/tex]
After insertion energy 0.0154128 J
Change in energy
[tex]\Delta E=U'-U\\\Rightarrow \Delta E=0.0154128-0.004056\\\Rightarrow \Delta E=0.0113568\ J[/tex]
There is an increase in energy.
The dielectric constant of air is about 1.00059 here the dielectric constant is 3.8 which is higher so the energy increases.
Answer:
Explanation:
Capacitance, C = 12 micro farad
voltage, V = 26 V
dielectric constant, K = 3.8
(a) The formula for the energy stored is
U = 1/2 CV²
U = 0.5 x 12 x 10^-6 x 26 x 26
U = 0.00406 J
(b) U' = 1/2 C'V²
U' = 0.5 x K x C x V² = K U
U' = 3.8 x 4.056 x 10^-3
U' = 0.0154 J
(c) The energy increases. As the value of dielectric constant increases so the energy increases.
