Answer:
Explanation:
The capacitance of the parallel-plate capacitor is given by
where
ϵ0 = 8.85x10-12 C2/N.m2 is the vacuum permittivity
k = 3.00 is the dielectric constant
is the area of the plates
d = 9.00 mm = 0.009 m is the separation between the plates
Substituting,
Now we can calculate the energy of the capacitor, given by:
where
C is the capacitance
V = 15.0 V is the potential difference
Substituting,
Answer:
Explanation:
plate separation = 2.3 x 10⁻³ m
capacity C₁ = ε A / d
= ε A / 2.3 x 10⁻³
C₂ = ε A / 1.15 x 10⁻³
=
a ) when charge remains constant
energy =
q is charge and C is capacity
energy stored initially E₁=
energy stored finally E₂ =
=
=
=
= 4.19 J
b )
In this case potential diff remains constant
energy of capacitor = 1/2 C V²
energy is proportional to capacity as V is constant .
= 16.76 .