NCERT · Class 12 · Physics · Electrostatic Potential and CapacitanceDerive an expression for the capacitance of a parallel plate capacitor when a dielectric slab of thickness $t$ and dielectric constant $K$ is introduced between the plates. Also, find the capacitance of a parallel plate capacitor of plate area $A$ and separation $d$, and calculate the equivalent capacitance when three capacitors of capacitances $C1$, $C2$, and $C3$ are connected in series.

Part 1: Capacitance of a Parallel Plate Capacitor\nA parallel plate capacitor consists of two large parallel conducting plates, each of area $A$, separated by a small distance $d$.

• Surface charge density on the plates: $\sigma = \frac{Q}{A}$
• Electric field between the plates in vacuum: $E_0 = \frac{\sigma}{\varepsilon_0} = \frac{Q}{A\varepsilon_0}$
• Potential difference between the plates: $V = E_0 d = \frac{Qd}{A\varepsilon_0}$
• Capacitance $C_0$ is given by: $$C_0 = \frac{Q}{V} = \frac{Q}{\frac{Qd}{A\varepsilon_0}} = \frac{\varepsilon_0 A}{d}$$

Part 2: Dielectric Slab Between the Plates\nLet a dielectric slab of thickness $t$ ($t < d$) and dielectric constant $K$ be introduced between the plates.

• Thickness of the region with air/vacuum = $(d - t)$
• Electric field in the dielectric medium: $E = \frac{E_0}{K}$
• Total potential difference $V$ between the plates is the sum of potentials across the vacuum region and the dielectric region: $$V = E_0(d - t) + E(t)$$\nSubstitute $E = \frac{E_0}{K}$ and $E_0 = \frac{Q}{A\varepsilon_0}$: $$V = E_0(d - t) + \frac{E_0}{K} t = E_0 \left( d - t + \frac{t}{K} \right)$$ $$V = \frac{Q}{A\varepsilon_0} \left( d - t + \frac{t}{K} \right)$$
• New capacitance $C$ with the dielectric slab: $$C = \frac{Q}{V} = \frac{Q}{\frac{Q}{A\varepsilon_0} \left( d - t + \frac{t}{K} \right)} = \frac{\varepsilon_0 A}{d - t + \frac{t}{K}}$$

Part 3: Equivalent Capacitance in Series Combination\nWhen three capacitors of capacitances $C_1$, $C_2$, and $C_3$ are connected in series:

• The charge $Q$ on each capacitor is the same.
• The total potential difference $V$ is the sum of potential differences across individual capacitors: $$V = V_1 + V_2 + V_3$$\nSince $V = \frac{Q}{C}$, we have: $$\frac{Q}{C_s} = \frac{Q}{C_1} + \frac{Q}{C_2} + \frac{Q}{C_3}$$\nDividing both sides by $Q$, the equivalent capacitance $C_s$ in series is: $$\frac{1}{C_s} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3}$$