Answer:
The energy stored in the capacitor after replacing the dielectric material = 2.00 J
Explanation:
The energy stored in a capacitor is given by
E = ½CV²
where C = capacitance of the capacitor; initially with vacuum as dielectric material is equal to 1.30 mF
V = voltage across the capacitor = 29.6 V
The capacitance of a capacitor is also given as
C = (Aε/d)
A = Cross sectional Area of the capacitor
d = distance between the plates of the capacitor
ε = permissivity of material between the plates of the capacitor
ε = kε₀
k = relative permissivity
ε₀ permissivity of vacuum.
If the Cross sectional Area of the capacitor and the distance between the plates of the capacitor are constant, it is evident that the capacitance of the capacitor is directly proportional to the permissivity of the dielectric material.
C₀ = (Aε₀/d) = 1.30 mF
C = (Aε/d) = (kAε₀/d) = kC₀
For the is question, the relative permissivity = 3.51
C = 3.51 × 1.30 mF = 4.563 mF = 0.004563 F
Energy stored in a capacitor is still
E = ½CV²
C = 4.563 mF = 0.004563 F
V = 29.6 V
E = (1/2)(0.004563)(29.6²)
E = 1.99895904 J = 2.00 J
Hope this Helps!!!
