Effect of Dielectric on Capacitance

The derivation CBSE sets

Two plates of area $A$ separated by $d$, carrying $\pm Q$, i.e. $\pm\sigma$ with $\sigma=\dfrac{Q}{A}$.

With vacuum between the plates:

$$E_0=\dfrac{\sigma}{\varepsilon_0},\qquad V_0=E_0d,\qquad C_0=\dfrac{Q}{V_0}=\dfrac{\varepsilon_0 A}{d}$$

Now fill the gap completely with a dielectric. It is polarised, and is equivalent to two sheets of bound charge $\pm\sigma_P$ on its faces normal to the field, so the effective density producing the field is $\sigma-\sigma_P$:

$$E=\dfrac{\sigma-\sigma_P}{\varepsilon_0},\qquad V=Ed=\dfrac{(\sigma-\sigma_P)d}{\varepsilon_0}$$

For a linear dielectric $\sigma_P\propto\sigma$, so $\sigma-\sigma_P=\dfrac{\sigma}{K}$ with $K$ a constant of the material and $K>1$. Then

$$V=\dfrac{\sigma d}{K\varepsilon_0}=\dfrac{Qd}{K\varepsilon_0 A},\qquad C=\dfrac{Q}{V}=\dfrac{K\varepsilon_0 A}{d}$$

$$\varepsilon=\varepsilon_0K,\qquad K=\dfrac{\varepsilon}{\varepsilon_0},\qquad C=KC_0$$

The dielectric constant is thus the factor, greater than one, by which the capacitance rises when the dielectric fills the gap. Though derived for parallel plates, $C=KC_0$ holds for any capacitor and can be taken as the general definition of $K$.

Equivalently, using $E=\dfrac{E_0}{K}=\dfrac{\sigma}{K\varepsilon_0}$ directly: $V=Ed=\dfrac{\sigma d}{K\varepsilon_0}$ and $C=\dfrac{Q}{V}=\dfrac{K\varepsilon_0A}{d}$ with $Q=\sigma A$.

Why the field falls when a slab is inserted

The dielectric is polarised; the bound surface charges produce a field opposing the applied field. With the plate charge fixed, $E_0=\dfrac{\sigma}{\varepsilon_0}$ and the net field becomes $E=\dfrac{E_0}{K}$, smaller by the factor $K$.

Partial filling

Slab of dielectric constant $K$ and thickness $t<d$, the rest of the gap empty:

$$C=\dfrac{\varepsilon_0 A}{d-t+\dfrac{t}{K}}$$

• A conducting slab is the limit $K\to\infty$: $C=\dfrac{\varepsilon_0 A}{d-t}$, and the result does not depend on where in the gap the slab sits.
• NCERT's worked case, $t=\dfrac{3d}{4}$: the potential falls by the factor $\dfrac{K+3}{4K}$ while the free charge is unchanged, so $C=\dfrac{4K}{K+3}C_0$.

Two slabs filling the gap

Stacked one above the other, each of thickness $d/2$ and full area $A$ — these are two capacitors in series:

$$C_1'=\dfrac{2K_1\varepsilon_0 A}{d},\qquad C_2'=\dfrac{2K_2\varepsilon_0 A}{d}$$

$$\dfrac{1}{C_s}=\dfrac{d}{2\varepsilon_0K_1A}+\dfrac{d}{2\varepsilon_0K_2A}=\left(\dfrac{K_1+K_2}{K_1K_2}\right)\dfrac{d}{2\varepsilon_0A}\ \Rightarrow\ C_s=\left(\dfrac{2K_1K_2}{K_1+K_2}\right)\dfrac{\varepsilon_0A}{d}$$

Setting this equal to a single slab of constant $K$ filling the gap gives the relation CBSE asks for:

$$K=\dfrac{2K_1K_2}{K_1+K_2}$$

Unequal thicknesses work the same way. Slabs of $2K$ over a thickness $d/3$ and $K$ over $2d/3$:

$$C_1=\dfrac{2K\varepsilon_0A}{d/3}=\dfrac{6K\varepsilon_0A}{d},\qquad C_2=\dfrac{K\varepsilon_0A}{2d/3}=\dfrac{3K\varepsilon_0A}{2d}$$

$$\dfrac{1}{C}=\dfrac{d}{6K\varepsilon_0A}+\dfrac{2d}{6K\varepsilon_0A}=\dfrac{3d}{6K\varepsilon_0A}\ \Rightarrow\ C=\dfrac{2K\varepsilon_0A}{d}$$

Side by side, each occupying half the area over the full separation $d$ — these are two capacitors in parallel:

$$C_p=\dfrac{K_1\varepsilon_0(A/2)}{d}+\dfrac{K_2\varepsilon_0(A/2)}{d}=\dfrac{\varepsilon_0A(K_1+K_2)}{2d}$$

• Rule of thumb: slabs stacked across the separation are in series, slabs placed side by side across the area are in parallel.

Battery connected or disconnected

This decides every "how is it affected" question. Insert a dielectric of constant $K$ into a capacitor of capacitance $C$:

• Worked pattern: $A=6\times10^{-3}\ \text{m}^{2}$, $d=3\ \text{mm}$ gives $C=17.7\ \text{pF}$; on $100\ \text{V}$, $Q=17.7\times10^{-10}\ \text{C}$. Insert mica with $K=6$ while the supply stays connected: $Q'=KQ=1.062\times10^{-8}\ \text{C}$, six times as much.
• Battery disconnected then $K$ inserted: $V'=\dfrac{Q}{KC}=\dfrac{V}{K}$ — the potential falls to $V/K$.

Separation changed with the battery disconnected

With $Q$ fixed and $d$ doubled: $C'=\dfrac{\varepsilon_0A}{2d}=\dfrac{C}{2}$, so the capacitance halves; but $E=\dfrac{\sigma}{\varepsilon_0}=\dfrac{Q}{\varepsilon_0A}$ does not contain $d$ at all, so the field between the plates is unchanged. The potential difference doubles and $U=\dfrac{Q^{2}}{2C}$ doubles. If a slab of constant $K$ with $1<K<2$ is then inserted to fill the doubled gap, $C''=\dfrac{K\varepsilon_0A}{2d}=\dfrac{K}{2}C$, the field becomes $\dfrac{E}{K}$, and

$$U''=\dfrac{Q^{2}}{2C''}=\dfrac{2}{K}\cdot\dfrac{Q^{2}}{2C}=\dfrac{2U}{K}>U\quad\text{since }1<K<2$$

so the stored energy increases.

Diagram to describe in words

Two plates with $+\sigma$ on the left and $-\sigma$ on the right. Between them a slab of thickness $t$ shaded, with a column of induced $-$ signs on its left face and $+$ signs on its right face, and thinner field arrows inside the slab than in the empty parts of the gap. For the two-slab cases, one figure with two horizontal shaded bands of thickness $d/2$ stacked between the plates (series), and one with two vertical shaded halves side by side spanning the full gap (parallel).

Standard results