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Capacitor Partially Filled With Dielectric

Capacitor Partially Filled With Dielectric :- Consider a parallel plate capacitor with plate area A and distance d between the plates. A dielectric slab of thickness t (t < d) and dielectric constant K is placed between the plates of the parallel plate capacitor as shown in the figure.

Let’s assume the plates have charges of ±Q, which creates an electric field E₀ in the air gap between the plates. Due to this electric field E₀, the molecules of the dielectric material become polarized, and an electric field E_p is induced in the dielectric material in the opposite direction to E₀. Therefore, the effective electric field in the dielectric material becomes E₀ – E_p, but the electric field in the air gap (d – t) remains E₀.

The potential difference (V) between the plates of the capacitor,

$\displaystyle V={{E}_{0}}(d-t)+Et$

But $\displaystyle \frac{{{E}_{0}}}{E}=K$

$\displaystyle \therefore E=\frac{{{E}_{0}}}{K}$

$\displaystyle \therefore V={{E}_{0}}(d-t)+\frac{{{E}_{0}}}{K}t$

$\displaystyle \Rightarrow V={{E}_{0}}\left[ d-t\text{+}\frac{t}{K} \right]$

The electric field E₀ in the air space between the capacitor plates,

$\displaystyle {{E}_{0}}=\frac{Q}{A{{\varepsilon }_{0}}}$

$\displaystyle \therefore V=\frac{Q}{A{{\varepsilon }_{0}}}\left[ d-t\text{+}\frac{t}{K} \right]$

Hence capacitance C ,

$\displaystyle C=\frac{Q}{V}=\frac{Q}{\frac{Q}{A{{\varepsilon }_{0}}}\left[ d-t\text{+}\frac{t}{K} \right]}=\frac{{{\varepsilon }_{0}}A}{d-t\text{+}\frac{t}{K}}$

$\displaystyle \therefore C=\frac{{{\varepsilon }_{0}}A}{d-t\left( 1-\frac{1}{K} \right)}$ …..(1)

Equation (1) is the expression for the capacitance of a parallel plate capacitor in the presence of a partially filled dielectric material.

Since the value of K is always greater than one, the value of $\displaystyle \left( 1-\frac{1}{K} \right)$ will always be positive. Thus, from equation (1), we can see that the distance d between the capacitor plates is effectively reduced by $\displaystyle t\left( 1-\frac{1}{K} \right)$ . As a result, the capacitance of the capacitor increases.

Note :-

(1). If the entire space between the plates is filled with a dielectric material, i.e., t = d , then,

$\displaystyle C=\frac{{{\varepsilon }_{0}}A}{d-d\left( 1-\frac{1}{K} \right)}=\frac{{{\varepsilon }_{0}}A}{d-d+\frac{d}{K}}=K\frac{{{\varepsilon }_{0}}A}{d}$ …..(2)

(2). If the entire space between the plates of a charged capacitor is filled with a dielectric material, and the battery remains connected, then :-

(i) The potential difference between the plates remains constant, that is,

$\displaystyle V={{V}_{0}}$

(ii) The capacitance increases,

$\displaystyle {{C}_{m}}=K{{C}_{0}}$

(iii) The charge on the plates increases,

$\displaystyle Q={{C}_{m}}V=(K{{C}_{0}})V=K\left( {{C}_{0}}V \right)=K{{Q}_{0}}$

(iv) The electric field intensity between the plates decreases,

$\displaystyle E=\frac{{{E}_{0}}}{K}$

(v) The energy stored in the capacitor increases,

$\displaystyle U=\frac{1}{2}{{C}_{m}}{{V}^{2}}=\frac{1}{2}{{\left( K{{C}_{0}} \right)}^{2}}{{V}^{2}}=K\left[ \frac{1}{2}{{C}_{0}}{{V}^{2}} \right]=K{{U}_{0}}$

(3). If a charged capacitor is disconnected from the battery and a dielectric material is filled in the entire space between the plates, then :-

(i) The potential difference between the plates decreases, that is,

$\displaystyle V=\frac{Q}{{{C}_{m}}}=\frac{Q}{K{{C}_{0}}}=\frac{1}{K}\left( \frac{Q}{{{C}_{0}}} \right)=\frac{{{V}_{0}}}{K}$

(ii) The capacitance increases,

$\displaystyle {{C}_{m}}=K{{C}_{0}}$

(iii) The magnitude of the charge on the plates remains the same, that is,

$\displaystyle Q={{Q}_{0}}$

(iv) The electric field intensity between the plates decreases,

$\displaystyle E=\frac{{{E}_{0}}}{K}$

(v) The energy stored in the capacitor decreases,

$\displaystyle U=\frac{1}{2}{{C}_{m}}{{V}^{2}}=\frac{1}{2}(K{{C}_{0}}){{\left( \frac{{{V}_{0}}}{K} \right)}^{2}}=\frac{1}{K}\left[ \frac{1}{2}{{C}_{0}}V_{0}^{2} \right]=\frac{{{U}_{0}}}{K}$

(4). If the entire space between the plates is filled with air, i.e., K = 1 and t = 0, then,

$\displaystyle C=\frac{{{\varepsilon }_{0}}A}{d-0\left( 1-\frac{1}{1} \right)}=\frac{{{\varepsilon }_{0}}A}{d}$ …..(3)

(5). If a metal plate of thickness t is placed partially between the plates (K = ∞), then the capacitance of a parallel plate capacitor,

In the above diagram, the effective electric field becomes zero in the metal plate, and the electric field E₀ exists only in the air gap (d – t).

Therefore, the potential difference (V) between the plates of the capacitor is,

$\displaystyle V={{E}_{0}}(d-t)=\frac{Q}{A{{\varepsilon }_{0}}}(d-t)$

Capacitance C ,

$\displaystyle C=\frac{Q}{V}=\frac{Q}{\frac{Q}{A{{\varepsilon }_{0}}}(d-t)}=\frac{{{\varepsilon }_{0}}A}{\left( d-t \right)}$ …..(4)

The above result can also be obtained by setting K = ∞ in equation (1) :-

$\displaystyle C=\frac{{{\varepsilon }_{0}}A}{d-t\left( 1-\frac{1}{\infty } \right)}=\frac{{{\varepsilon }_{0}}A}{\left( d-t \right)}$

Here, the distance d between the capacitor plates effectively decreases to (d – t), therefore, placing a metal plate partially between the capacitor plates increases the capacitance.

(6). If the entire space between the plates is filled with metal, i.e., t = d, then,

$\displaystyle C=\frac{{{\varepsilon }_{0}}A}{\left( d-d \right)}=\infty$

(7). If multiple dielectric slabs with dielectric constants K₁, K₂, K₃,…… and thicknesses t₁, t₂, t₃,…… respectively, are placed between the plates, then the capacitance of the capacitor is given by :

$\displaystyle C=\frac{{{\varepsilon }_{0}}A}{d-{{t}_{1}}\left( 1-\frac{1}{{{K}_{1}}} \right)-{{t}_{2}}\left( 1-\frac{1}{{{K}_{2}}} \right)-{{t}_{3}}\left( 1-\frac{1}{{{K}_{3}}} \right)-.....}$

$\displaystyle \Rightarrow C=\frac{{{\varepsilon }_{0}}A}{\left[ d-\left( {{t}_{1}}+{{t}_{2}}+{{t}_{3}}+..... \right)+\frac{{{t}_{1}}}{{{K}_{1}}}+\frac{{{t}_{2}}}{{{K}_{2}}}+\frac{{{t}_{3}}}{{{K}_{3}}}+..... \right]}$

But $\displaystyle \left( {{t}_{1}}+{{t}_{2}}+{{t}_{3}}+..... \right)=d$ , Therefore, capacitance

$\displaystyle C=\frac{{{\varepsilon }_{0}}A}{\left[ \frac{{{t}_{1}}}{{{K}_{1}}}+\frac{{{t}_{2}}}{{{K}_{2}}}+\frac{{{t}_{3}}}{{{K}_{3}}}+..... \right]}$ …..(5)

Capacitor Partially Filled With Dielectric

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