Question

When an air capacitor with a capacitance of 380 nF (1 {\rm nF} = 10^{-9}\;{\rm F}) is connected to a power supply, the energy stored in the capacitor is 1.95×10−5 J. While the capacitor is kept connected to the power supply, a slab of dielectric is inserted that completely fills the space between the plates. This increases the stored energy by 2.62×10−5 J.
A) What is the potential difference between the capacitor plates?
B)What is the dielectric constant of the slab?

Answer 1

Stored Energy = (1/2)CV^2
It increases by a factor (2.62+1.95)/1.95 = 2.344
A) Since V is constant, C increases by the same factor, to
2.344*380 nF = 891 nF
B) The dielectric constant is the factor that increased the capacitance, 2.344

Answer 2

Wgat is energy

Answer 3

Nonsense.