1)
The capacitance of a parallel-plate capacitor is given by:
where
is the vacuum permittivity
A is the area of each plate
d is the distance between the plates
Here, the radius of each plate is
so the area is
While the separation between the plates is
So the capacitance is
And now we can find the energy stored,which is given by:
2) 0.71 J/m^3
The magnitude of the electric field is given by
and the energy density of the electric field is given by
and using
, we find
Answer:
the coefficient of volume expansion of the glass is
Explanation:
Given that:
Initial volume of the glass flask = 1000 cm³ = 10⁻³ m³
temperature of the glass flask and mercury= 1.00° C
After heat is applied ; the final temperature = 52.00° C
Temperature change ΔT = 52.00° C - 1.00° C = 51.00° C
Volume of the mercury overflow = 8.50 cm^3 = 8.50 × 10⁻⁶ m³
the coefficient of volume expansion of mercury is 1.80 × 10⁻⁴ / K
The increase in the volume of the mercury = 10⁻³ m³ × 51.00 × 1.80 × 10⁻⁴
The increase in the volume of the mercury =
Increase in volume of the glass = 10⁻³ × 51.00 ×
Now; the mercury overflow = Increase in volume of the mercury - increase in the volume of the flask
the mercury overflow =
Thus; the coefficient of volume expansion of the glass is