Capacitor and Capacitance — Revision Notes

⚡ 30-Second Revision

Capacitance Definition: — $C = Q/V$
Unit of Capacitance: — Farad (F) = Coulomb/Volt
Parallel Plate Capacitor (Vacuum/Air): — $C = \frac{A\epsilon_0}{d}$
Parallel Plate Capacitor (Dielectric): — $C = \frac{\kappa A\epsilon_0}{d} = \kappa C_{\text{air}}$
Energy Stored: — $U = \frac{1}{2}CV^2 = \frac{1}{2}QV = \frac{Q^2}{2C}$
Electric Field (Parallel Plate): — $E = V/d = \frac{\sigma}{\epsilon_0} = \frac{Q}{A\epsilon_0}$
Effect of Dielectric (Disconnected): — $Q$ constant, $V \downarrow$, $E \downarrow$, $C \uparrow$
Effect of Dielectric (Connected): — $V$ constant, $Q \uparrow$, $E$ constant, $C \uparrow$

2-Minute Revision

Capacitors are devices that store electrical energy in an electric field. They consist of two conductors separated by an insulator called a dielectric. The ability of a capacitor to store charge is quantified by its capacitance ($C$), defined as the ratio of charge ($Q$) on one plate to the potential difference ($V$) across the plates: $C = Q/V$.

The SI unit is the Farad (F). For a parallel plate capacitor, capacitance depends on the plate area ($A$), separation ($d$), and the dielectric constant ($\kappa$) of the material between plates: $C = \frac{\kappa A\epsilon_0}{d}$.

A larger area and smaller separation lead to higher capacitance. Inserting a dielectric material (with $\kappa > 1$) always increases capacitance. The energy stored in a capacitor is given by $U = \frac{1}{2}CV^2$.

When a capacitor is disconnected from a battery, its charge $Q$ remains constant. If a dielectric is then inserted, $V$ decreases and $C$ increases. If it remains connected to the battery, $V$ remains constant, and $Q$ increases as $C$ increases.

Remember to convert units to SI before calculations.

5-Minute Revision

Capacitors are fundamental components for storing electrical energy. Their core principle involves separating positive and negative charges on two conductive plates, creating an electric field in the insulating dielectric material between them.

This charge separation results in a potential difference across the plates. Capacitance ($C$) is the measure of this charge-storing ability, defined as $C = Q/V$, where $Q$ is the charge magnitude on one plate and $V$ is the potential difference.

The unit is the Farad (F), though microfarads ($\mu\text{F}$) and picofarads ($\text{pF}$) are more common. For a parallel plate capacitor, the capacitance is $C = \frac{A\epsilon_0}{d}$ in vacuum/air, where $A$ is plate area and $d$ is separation.

If a dielectric material with dielectric constant $\kappa$ is inserted, capacitance increases to $C = \frac{\kappa A\epsilon_0}{d}$. This increase is due to the dielectric's polarization, which reduces the net electric field and thus the potential difference for a given charge.

The energy stored in a capacitor is $U = \frac{1}{2}CV^2 = \frac{1}{2}QV = \frac{Q^2}{2C}$. It's crucial to understand the behavior when a dielectric is inserted: if the capacitor is disconnected from the source, $Q$ is constant, so $V$ decreases and $C$ increases.

If it remains connected, $V$ is constant, so $Q$ increases and $C$ increases. Always convert units to SI (e.g., $\mu\text{F}$ to F, cm to m) for calculations. Practice problems involving these formulas and conceptual scenarios, especially those involving changes in geometry or dielectric insertion, are key to mastering this topic for NEET.

Mini-Example: A $5,mu\text{F}$ capacitor is charged to $50,\text{V}$. What is the charge stored and energy stored?

Solution:

1
Convert $C = 5,mu\text{F} = 5 \times 10^{-6},\text{F}$. $V = 50,\text{V}$.
2
Charge stored: $Q = CV = (5 \times 10^{-6},\text{F}) \times (50,\text{V}) = 250 \times 10^{-6},\text{C} = 2.5 \times 10^{-4},\text{C}$.
3
Energy stored: $U = \frac{1}{2}CV^2 = \frac{1}{2} \times (5 \times 10^{-6},\text{F}) \times (50,\text{V})^2 = \frac{1}{2} \times 5 \times 10^{-6} \times 2500,\text{J} = 6250 \times 10^{-6},\text{J} = 6.25 \times 10^{-3},\text{J}$.

Prelims Revision Notes

1
Capacitor Definition: — Device storing electrical energy in an electric field. Two conductors (plates) separated by a dielectric (insulator).
2
Capacitance (C): — Ability to store charge. $C = Q/V$. SI unit: Farad (F). $1,\text{F} = 1,\text{C/V}$. Common units: $\mu\text{F} (10^{-6},\text{F})$, $\text{nF} (10^{-9},\text{F})$, $\text{pF} (10^{-12},\text{F})$.
3
Factors Affecting Capacitance:

* Geometry: Plate area ($A$) and separation ($d$). * Dielectric Material: Dielectric constant ($\kappa$).

1
Parallel Plate Capacitor:

* Vacuum/Air: $C = \frac{A\epsilon_0}{d}$. ($\epsilon_0 = 8.85 \times 10^{-12},\text{F/m}$) * With Dielectric: $C = \frac{\kappa A\epsilon_0}{d} = \kappa C_{\text{air}}$. ($\kappa \ge 1$)

1
Energy Stored in a Capacitor:

* $U = \frac{1}{2}CV^2$ * $U = \frac{1}{2}QV$ * $U = \frac{Q^2}{2C}$

1
Electric Field between Plates: — $E = V/d$. Also, $E = \sigma/\epsilon_0 = Q/(A\epsilon_0)$.
2
Effect of Dielectric Insertion:

* Capacitor Disconnected from Battery: Charge ($Q$) remains constant. Capacitance ($C$) increases by factor $\kappa$. Potential difference ($V$) decreases by factor $\kappa$ ($V' = V/\kappa$). Electric field ($E$) decreases by factor $\kappa$ ($E' = E/\kappa$).

Energy stored ($U$) decreases by factor $\kappa$ ($U' = U/\kappa$). * Capacitor Connected to Battery: Potential difference ($V$) remains constant. Capacitance ($C$) increases by factor $\kappa$. Charge ($Q$) increases by factor $\kappa$ ($Q' = \kappa Q$).

Electric field ($E$) remains constant. Energy stored ($U$) increases by factor $\kappa$ ($U' = \kappa U$).

1
Important Note: — Capacitance is an intrinsic property of the capacitor's physical structure and dielectric. It does NOT depend on the charge stored or the voltage applied. $C=Q/V$ is a definition, not a variable relationship.

Vyyuha Quick Recall

Can Always Decrease Energy Keeping Quietly Voltage. (C = A * epsilon_0 / d, E = V/d, Q = CV, K = dielectric constant). This mnemonic helps recall the parallel plate capacitor formula and the relationships between C, Q, V, E, and the dielectric constant. Or, for the effect of dielectric: Disconnected Quietly Vanishes Energy, Connected Voltage Quickly Expands.