Answer:
U = 218 nJ
Explanation:
We are given;
Spacing between the plates; d = 3.4 mm = 3.4 × 10^(-3) m
Voltage across the capacitor; V = 96 V
Dimension of the square plates is 7.2cm x 7.2cm.
So, Area = 7.2 × 7.2 = 51.84 cm² = 51.84 × 10^(-4) m²
Permittivity of free space; ε_o = 8.85 × 10^(-12) C²/N.m²
From relative permeability table;
Dielectric constant of Pyrex; k1 = 5.6
Dielectric constant of polystyrene; k2 = 2.56
Now, formula for capacitance of a capacitor with Dielectric is;
C = kC_o
Where, C_o = ε_o(A/d)
Since there are 2 capacitors, d will now be d/2 = (3.4 × 10^(-3))/2 m = 1.7 × 10^(-3)
Since we have 2 capacitor, thus ;
C1 = k1*ε_o*(A/d)
C1 = (5.6 × 8.85 × 10^(-12) × (51.84 × 10^(-4))/(1.7 × 10^(-3))
C1 = 1.51 × 10^(-10) F
Similarly;
C2 = (2.56 × 8.85 × 10^(-12) × (51.84 × 10^(-4))/(1.7 × 10^(-3))
C2 = 0.691 × 10^(-10) F
For capacitors in series, formula for total capacitance(Cs) is;
1/Cs = (1/C1) + (1/C2)
Simplifying this, we have;
Cs = (C1*C2)/(C1 + C2)
Plugging in the relevant values ;
Cs = (1.51 × 10^(-10)*0.691 × 10^(-10))/((1.51 × 10^(-10)) + (0.691 × 10^(-10)))
Cs = 0.474 × 10^(-10) F
The formula for energy stored in a capacitor with 2 Dielectrics is given as;
U = ½Cs*V²
So,
U = ½ × 0.474 × 10^(-10) × 96²
U = 2.18 × 10^(-7) J = 218 × 10^(-9) = 218 nJ
rad/s
--- (1)
rad/s