Jee Mains Capacitance Study Notes: Physics Wallah Cheatsheet

Fundamentals of Capacitance

• Definition and Basic Formula: The physical property of a system that describes its ability to store electric charge ($Q$) per unit of electric potential ($V$) is Capacitance. It is governed by the equation:     $Q = CV$$\text{Alternatively, } C = \frac{Q}{V}$
• Parallel Plate Capacitor Formula: The capacitance for a parallel plate setup in a vacuum or air is given by:     $C = \frac{\varepsilon_0 A}{d}$
• Factors Affecting Capacitance: Capacitance depends on the following three physical and environmental factors:
  1. Distance between plates ($d$): The closer the plates, the higher the capacitance.
  2. Area of plates ($A$): A larger surface area allows for more charge storage.
  3. Medium between plates: The presence of a dielectric material increases capacitance.
• Factors NOT Affecting Capacitance: Capacitance is an intrinsic property of the physical arrangement and is independent of:
  1. Charge ($Q$): Increasing the charge on a capacitor increases its potential but not its capacitance.
  2. Potential difference ($V$).
• Unit of Capacitance: The standard International System of Units (SI) unit for capacitance is the Farad ($F$).

Energy Stored in a Capacitor

• Work Done by Battery: When a battery moves a total charge ($Q$) through a potential difference ($V$), the total work performed is:     $\text{Work done} = QV$
• Energy Distribution: During the charging process, energy is split between storage and dissipation:
  1. Stored Energy ($50\%$): The energy successfully stored in the electric field of the capacitor is denoted as $U$.         $U = \frac{1}{2} QV = \frac{1}{2} CV^2 = \frac{Q^2}{2C}$
  2. Dissipated Energy ($50\%$): Half of the energy provided by the battery is lost as heat ($H$) due to resistance in the connecting wires or the charging process.         $\text{Heat loss} = \frac{1}{2} QV = \frac{1}{2} CV^2$

Parallel Plate Capacitor and Separation Variations

• Capacitance Relation: $C = \frac{\varepsilon_0 A}{d}$ implies $C \propto \frac{1}{d}$.
• Electric Field ($E$): The field between the plates is uniform and defined by:     $E = \frac{\sigma}{\varepsilon_0} = \frac{Q}{A \varepsilon_0}$$E = \frac{V}{d}$
• Potential ($V$): $V = Ed = \frac{Qd}{A \varepsilon_0}$.
• Force between plates of a parallel plate capacitor: Each plate exerts an attractive force on the other, calculated as:     $F = Q \times \text{Field due to one plate} = Q \times \frac{\sigma}{2 \varepsilon_0} = Q \times \frac{Q}{2A \varepsilon_0}$$F = \frac{Q^2}{2A \varepsilon_0}$
• Energy ($U$): $U = \frac{1}{2} CV^2$ or $U = \frac{Q^2 d}{2A \varepsilon_0}$.

Changing Plate Separation: Battery Conditions

When the plate separation ($d$) is changed (e.g., doubled from $d$ to $2d$), the results depend on whether the battery remains connected or is disconnected.

Battery Remains Connected

In this case, the potential difference ($V$) across the capacitor remains constant because it is determined by the battery voltage.

1. Voltage: $V' = \text{Constant}$
2. Capacitance: $C' = \frac{C}{2}$ ($C \propto \frac{1}{2d}$)
3. Charge: $Q' = \frac{Q}{2}$ (Since $Q = CV$ and $C$ is halved)
4. Electric Field: $E' = \frac{E}{2}$ ($E = \frac{V}{d}$)
5. Force: $F' = \frac{F}{4}$ ($F = \frac{Q^2}{2A \varepsilon_0}$, $Q^2$ is reduced by 4)
6. Energy: $U' = \frac{U}{2}$ ($U = \frac{1}{2} CV^2$

Battery Disconnected

In this case, the charge ($Q$) remains constant because there is no path for the charge to leave or enter the plates.

1. Charge: $Q' = \text{Constant}$
2. Capacitance: $C' = \frac{C}{2}$
3. Voltage: $V' = 2V$ (Since $V = \frac{Q}{C}$ and $C$ is halved)
4. Electric Field: $E' = E$ ($\text{Constant}$ because $Q$, $A$, and $\varepsilon_0$ have not changed)
5. Force: $F' = F$ ($\text{Constant}$ because $Q$ depends on plates)
6. Energy: $U' = 2U$ ($U = \frac{Q^2}{2C}$, $C$ is halved)

Dielectric in a Capacitor

• Effect on Capacitance: Inserting a dielectric with dielectric constant ($K$) increases the capacitance.     $C' = KC$$C' = \frac{K \varepsilon_0 A}{d}$

Impact of Dielectric Insertion

Connecting Two Charged Capacitors

When two capacitors ($C_1$ with potential $V_1$ and $C_2$ with potential $V_2$) are connected together, charge redistributes until they reach a common potential.

Case 1: Same Polarity Connected Together

Plates with similar charges ($+$ to $+$) are connected.

• Common Potential: $V_{\text{common}} = \frac{C_1 V_1 + C_2 V_2}{C_1 + C_2}$
• Energy/Heat Loss: $\Delta U = \frac{C_1 C_2 (V_1 - V_2)^2}{2(C_1 + C_2)}$

Case 2: Opposite Polarity Connected Together

Plates with opposite charges ($+$ to $-$) are connected.

• Common Potential: $V_{\text{common}} = \frac{C_1 V_1 - C_2 V_2}{C_1 + C_2}$
• Energy/Heat Loss: $\Delta U = \frac{C_1 C_2 (V_1 + V_2)^2}{2(C_1 + C_2)}$

Grouping of Capacitors

1. Series Combination

Capacitors are connected end-to-end such that the charge ($Q$) is same on all capacitors.

• Equivalent Capacitance: $\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} + \dots + \frac{1}{C_n}$
• Potential Split: The total voltage ($V$) is the sum of individual voltages ($V = V_1 + V_2$).
• Voltage Divider Rule: The voltage across a capacitor is inversely proportional to its capacitance ($V \propto \frac{1}{C}$).     $V_1 = \left( \frac{C_2}{C_1 + C_2} \right) V$$V_2 = \left( \frac{C_1}{C_1 + C_2} \right) V$

2. Parallel Combination

Capacitors are connected between the same two points such that the potential difference ($V$) is the same across all capacitors.

• Equivalent Capacitance: $C_{eq} = C_1 + C_2 + \dots + C_n$
• Identical Capacitors: If $C_1 = C_2 = \dots = C_n = C$, then $C_{eq} = nC$.
• Charge Split: The total charge ($Q$) is the sum of individual charges.
• Charge Divider Rule: Charge is directly proportional to capacitance ($Q \propto C$).     $Q_1 = \left( \frac{C_1}{C_1 + C_2} \right) Q$$Q_2 = \left( \frac{C_2}{C_1 + C_2} \right) Q$

Multiple Dielectrics

1. Series Combination (Stacked Slabs)

Dielectrics are stacked one after another between the plates (affecting thickness $d$).

• Equivalent Dielectric Constant: $K_{eq} = \frac{\sum d_i}{\sum \frac{d_i}{K_i}}$
• Case for two slabs of equal thickness ($d_1 = d_2$):     $K_{eq} = \frac{2K_1 K_2}{K_1 + K_2}$

2. Parallel Combination (Side-by-Side Slabs)

Dielectrics are placed next to each other (affecting Area $A$).

• Equivalent Dielectric Constant: $K_{eq} = \frac{\sum K_i A_i}{\sum A_i}$
• Case for two slabs of equal area ($A_1 = A_2$):     $K_{eq} = \frac{K_1 + K_2}{2}$

Spherical Capacitor and Charge Redistribution

Spherical Capacitor

• Isolated Sphere: $C = 4 \pi \varepsilon_0 R$
• Concentric Spheres: $C = 4 \pi \varepsilon_0 \left( \frac{R_1 R_2}{R_2 - R_1} \right)$

Redistribution: The Big Drop Problem

When $n$ small drops (each with radius $r$, charge $q$, capacitance $c$, and potential $v$) coalesce to form a single big drop:

1. Charge: $Q = nq$
2. Capacitance: $C = n^{1/3} c$
3. Potential: $V = n^{2/3} v$

Wheatstone's Bridge

In a circuit with five capacitors arranged in a bridge shape, the bridge is balanced if the ratios of capacitances are equal:

• Condition: If $\frac{C_1}{C_2} = \frac{C_3}{C_4}$, then the bridge is balanced.
• Implied Property: $\text{Potential at point A (} V_A \text{) } = \text{ Potential at point B (} V_B \text{)}$.
• Consequence: No charge flows through the middle capacitor connecting A and B; it can be removed from calculation.