Question

When an air capacitor with a capacitance of 380 nf (1 nf = 10−9f) is connected to a power supply, the energy stored in the capacitor is 1.50×10−5 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 2.48×10−5 j?

Answer 1

Electrical potential energy stored in the capacitor U = 1/2* C V^2.
V = voltage of battery.
Capacitance C = K * C0
K = dielectric constant.
C0 = capacitance with air in the capacitor
V^2 = 2U/C0 = 3000/38
V = 8.88 Volts

Dielectric constant K = C/C0 = U / U0
K = (1.50+2.48)/1.50 = 2.65

Answer 2

Hey there !!!!!!!!

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Capacitance of a capacitor  = k*ξ₀*A/d

k= dielectric constant  ξ₀=Permittivity of free space  A= area occupied by plates  d=distance between two plates

 Energy (U) = Capacitance*(Potential)²/2 = C*V²/2

According to question potential between plates remains constant 

So Energy is directly proportional to Capacitance

Energy with a capacitor of capacitance 380nf present in air =1.5*10⁻⁵ J

Energy when a dielectric is inserted between the plates = 2.48*10⁻⁵J+1.5*10⁻⁵J = 3.98 *10⁻⁵J 

Capacitance of capacitor changes by "k" times  C₀ =k*C

U/U₀ = C/C₀

1.5*10⁻⁵/3.98*10⁻⁵ = C/k*C

1.5/3.98 = 1/k

k= 3.98/1.5

k =2.653.

Dielectric constant is a dimensionless quantity 

∴ k=2.653

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Hope this helped you..............