Question

Consider a air-filled parallel-plate capacitor with plates of length 8 cm, width 5.52 cm, spaced a distance 1.99 cm apart. Now imagine that a dielectric slab with dielectric constant 2.6 is inserted a length 4.4 cm into the capacitor. The slab has the same width as the plates. The capacitor is completely filled with dielectric material down to a length of 4.4 cm. A battery is connected to the plates so that they are at a constant potential 0.8 V while the dielectric is inserted.
Required:
What is the ratio of the new potential energy to the potential energy before the insertion of the dielectric?

Answer 1

Answer:

The ratio of the new potential energy to the potential energy before the insertion of the dielectric is 0.58

Explanation:

Given that,

Length of plates = 8 cm

Width = 5.52 cm

Distance = 1.99 cm

Dielectric constant = 2.6

Length = 4.4 cm

Potential = 0.8 V

We need to calculate the initial capacitance

Using formula of capacitance

$C=\dfrac{\epsilon_{0}A}{d}$

Put the value into the formula

$C=\dfrac{8.85\times10^{-12}\times8\times5.52\times10^{-4}}{1.99\times10^{-2}}$

$C=1.96\times10^{-12}$

We need to calculate the final capacitance

Using formula of capacitance

$C'=\dfrac{\epsilon_{0}A_{1}}{d}+\dfrac{k\epsilon_{0}A_{2}}{d}$

Put the value into the formula

$C'=(\dfrac{8.85\times10^{-12}}{1.99\times10^{-2}})((4.4\times5.52)+(3.6\times5.52)2.6)\times10^{-4}$

$C'=3.37\times10^{-12}$

We need to calculate the  ratio of the new potential energy to the potential energy before the insertion of the dielectric

Using formula of energy

$\dfrac{E}{E'}=\dfrac{\dfrac{1}{2}CV^2}{\dfrac{1}{2}C'V^2}$

Put the value into the formula

$\dfrac{E}{E'}=\dfrac{1.96\times10^{-12}}{3.37\times10^{-12}}$

$\dfrac{E}{E'}=0.58$

Hence, The ratio of the new potential energy to the potential energy before the insertion of the dielectric is 0.58