Question

A parallel-plate capacitor of plate separation d is charged to a potential difference V0. A dielectric slab of thickness d and dielectric constant k is introduced between the plates while the battery remains connected to the plates.

(a) show that the ratio of energy stored after the dielectic is introduced to the energy stored in the empty capacitor is U/U0 = k.

(b) Give a physical explanation for this increase in stored energy.

(c) What happens to the charge on the capacitor? Note: This situation is not the same as when the battery is removed from the circuit before the dielectric is introduced.

Answer 1

Answer:

Explanation:

(1.a)

Write the expression for old parallel plate capacitor.

$C = \frac{{{\varepsilon _0}A}}{d}$

The expression of energy stored in the capacitor without any dielectric between the plate is:

$U = \frac{1}{2}C{\left( {\Delta {V_0}} \right)^2}$

Substitute $\frac{{{\varepsilon _0}A}}{d}$ for C

$U = \frac{1}{2}\left( {\frac{{{\varepsilon _0}A}}{d}} \right){\left( {\Delta {V_0}} \right)^2}...(1)$

Supply source is same therefore, the potential difference is constant. So, the energy stored is:

${U_0} = \frac{1}{2}C'{\left( {\Delta {V_0}} \right)^2}$

Write the expression for new parallel plate capacitor.

$C' = \frac{{k{\varepsilon _0}A}}{d}$

Substitute $\frac{{k{\varepsilon _0}A}}{d}$ for C' in expression ${U_0}$

${U_0} = \frac{1}{2}\left( {\frac{{k{\varepsilon _0}A}}{d}} \right){\left( {\Delta {V_0}} \right)^2}...(2)$

From equation (1) and (2).

$\begin{array}{c}\\\frac{U}{{{U_0}}} = \frac{{\left( {\frac{1}{2}\left( {\frac{{{\varepsilon _0}A}}{d}} \right){{\left( {\Delta {V_0}} \right)}^2}} \right)}}{{\left( {\frac{1}{2}\left( {\frac{{k{\varepsilon _0}A}}{d}} \right){{\left( {\Delta {V_0}} \right)}^2}} \right)}}\\\\ = k\\\end{array}$

The ratio of energy stored after the dielectric is introduced to the energy stored in the empty capacitor is $\frac{U}{{{U_0}}} = k$

(1.b)

Write the expression for the capacitor$\left( C \right)$ .

$C = \frac{{k{\varepsilon _0}A}}{d}$

The expression for the energy stored by the capacitor is:

${U_0} = \frac{1}{2}C{\left( {\Delta {V_0}} \right)^2}$

Substitute $\frac{{k{\varepsilon _0}A}}{d}$ for C.

${U_0} = \frac{1}{2}\left( {\frac{{k{\varepsilon _0}A}}{d}} \right){\left( {\Delta {V_0}} \right)^2}$

When the distance between the plates is increased and the potential difference between the plates remains constant, then energy stored by the capacitor decreases. Similarly, if the distance between the plates decreases and potential remains constant, then energy stored by the capacitor is increased.

(1.c)

Write the expression for the original charge.

${Q_0} = {C_0}\Delta {V_0}...(3)$

After inserting the dielectric k , the new charge:

$Q = k{C_0}\Delta {V_0}...(4)$

From equation (3) and (4).

$\begin{array}{c}\\\frac{Q}{{{Q_0}}} = \frac{{k{C_0}\Delta {V_0}}}{{{C_0}\Delta {V_0}}}\\\\ = k\\\end{array}$

The charge on the capacitor increases by k