Electrostatics and Capacitors

🟢 Lite — Quick Review (1h–1d)

Rapid summary of electrostatics and capacitors for NABTEB physics.

Electrostatics is the study of stationary (static) electric charges and their interactions.

Fundamental Concepts:

• Charge ($q$): Measured in Coulombs (C). Two types: positive ($+$) and negative ($-$). Electrons carry negative charge; protons carry positive charge.
• Law of Charges: Like charges repel; unlike charges attract
• Conservation of Charge: Charge cannot be created or destroyed, only transferred
• Quantisation of Charge: Charge exists in discrete amounts: $q = ne$, where $n$ is an integer and $e = 1.6 \times 10^{-19}$ C

Coulomb’s Law: The force between two point charges: $$F = k\frac{q_1 q_2}{r^2}$$

Where $k = \frac{1}{4\pi\varepsilon_0} = 8.99 \times 10^9 , \text{N·m}^2/\text{C}^2$ (approximately $9 \times 10^9$)

The force is repulsive if $q_1 q_2 > 0$ (like charges) and attractive if $q_1 q_2 < 0$ (unlike charges).

Electric Field: The electric field ($E$) at a point is the force per unit charge: $$E = \frac{F}{q} = k\frac{Q}{r^2}$$

Direction: Away from positive charge; towards negative charge. Unit: N/C or V/m

Electric Potential: $$V = k\frac{Q}{r}$$

Unit: Volt (V). 1 V = 1 J/C.

Capacitance: $$C = \frac{Q}{V}$$

Unit: Farad (F). A capacitor stores charge when voltage is applied.

⚡ NABTEB Exam Tip: Electric field lines go FROM positive charges TO negative charges (externally). Inside a closed conducting shell, the field is zero (electrostatic shielding). Capacitors store energy: $U = \frac{1}{2}CV^2 = \frac{1}{2}QV^2 = \frac{Q^2}{2C}$.

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🟡 Standard — Regular Study (2d–2mo)

For NABTEB students who want a solid understanding of electrostatics.

Electric Field Patterns:

Charge Configuration	Field Pattern
Single positive charge	Radial lines outward
Single negative charge	Radial lines inward
Two like charges	Field lines repel each other
Two unlike charges	Field lines connect from + to −
Uniform field (parallel plates)	Parallel, equally spaced lines

Capacitors:

A capacitor consists of two conducting plates separated by an insulating material (dielectric).

Parallel Plate Capacitor: $$C = \frac{\varepsilon_0 \varepsilon_r A}{d}$$

Where $A$ = area of one plate, $d$ = separation between plates, $\varepsilon_r$ = relative permittivity of dielectric, $\varepsilon_0 = 8.85 \times 10^{-12}$ F/m.

Dielectrics: When an insulating material is placed between capacitor plates:

• It reduces the effective electric field
• Capacitance increases by factor $\varepsilon_r$ (dielectric constant)
• Different materials have different $\varepsilon_r$ values

Material	Dielectric Constant ($\varepsilon_r$)
Vacuum	1.00
Air	1.0006
Paper	3.5
Glass	5–10
Water	80

Energy Stored in a Capacitor: $$U = \frac{1}{2}CV^2 = \frac{1}{2}QV^2 = \frac{Q^2}{2C}$$

Capacitor Combinations:

Series Combination: $$\frac{1}{C_{\text{eq}}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3}$$

Charge is the same on all capacitors in series: $Q_1 = Q_2 = Q_3$ Voltage divides: $V = V_1 + V_2 + V_3$

Parallel Combination: $$C_{\text{eq}} = C_1 + C_2 + C_3$$

Voltage is the same across all capacitors in parallel: $V_1 = V_2 = V_3$ Charge divides: $Q = Q_1 + Q_2 + Q_3$

Applications of Capacitors:

1. Energy storage — in camera flashes, defibrillators
2. Smoothing — in power supplies to reduce voltage ripples
3. Tuning — in radio circuits (with inductors)
4. Timing — in RC circuits

Common Exam Mistakes:

• Confusing electric field ($E = F/q$) with electric potential ($V = kQ/r$)
• Forgetting that capacitors in series share charge (same $Q$)
• Not converting units consistently (e.g., using pF instead of F)
• Confusing direction of field lines (always from + to − externally)

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🔴 Extended — Deep Study (3mo+)

Comprehensive coverage of electrostatics and capacitors for NABTEB students.

Derivation of Coulomb’s Law:

Experimental observation shows that the force $F$ between two charges is:

• Proportional to each charge: $F \propto q_1$ and $F \propto q_2$
• Inversely proportional to square of distance: $F \propto 1/r^2$

Combining: $F = k\frac{q_1 q_2}{r^2}$

The constant $k = \frac{1}{4\pi\varepsilon_0}$ where $\varepsilon_0 = 8.85 \times 10^{-12}$ F/m is the permittivity of free space.

Vector Form of Coulomb’s Law: $$\vec{F}{12} = k\frac{q_1 q_2}{r{12}^2} \hat{r}_{12}$$

Where $\hat{r}_{12}$ is the unit vector from charge 1 to charge 2.

Superposition Principle:

For multiple point charges, the net force on any one charge is the vector sum of forces exerted by all other charges: $$\vec{F}{\text{net}} = \vec{F}{12} + \vec{F}{13} + \vec{F}{14} + \ldots$$

Similarly for electric field and potential: $$\vec{E}_{\text{net}} = \vec{E}_1 + \vec{E}_2 + \vec{E}3 + \ldots$$ $$V{\text{net}} = V_1 + V_2 + V_3 + \ldots \text{ (algebraic sum)}$$

Electric Field due to Common Charge Distributions:

1. Uniformly charged thin spherical shell:

• Inside the shell ($r < R$): $E = 0$
• On the surface ($r = R$): $E = k\frac{Q}{R^2}$
• Outside the shell ($r > R$): $E = k\frac{Q}{r^2}$ (treat as point charge)

2. Uniformly charged solid sphere (non-conducting):

• Inside ($r < R$): $E = k\frac{Qr}{R^3}$ (linearly increasing)
• Outside ($r > R$): $E = k\frac{Q}{r^2}$

3. Infinite plane sheet of charge: $$E = \frac{\sigma}{2\varepsilon_0}$$

Where $\sigma = Q/A$ is the surface charge density.

4. Two parallel oppositely charged plates: $$E = \frac{\sigma}{\varepsilon_0}$$

(Uniform field between plates, zero outside)

Electric Potential Energy:

The work done in bringing a charge $q$ from infinity to a point at distance $r$ from charge $Q$: $$U = k\frac{qQ}{r}$$

For multiple charges: $$U_{\text{total}} = k\left(\frac{q_1 q_2}{r_{12}} + \frac{q_1 q_3}{r_{13}} + \frac{q_2 q_3}{r_{23}} + \ldots\right)$$

Relationship between $E$ and $V$: $$E = -\frac{dV}{dr}$$

In a uniform field: $E = \frac{V}{d}$ where $d$ is the separation.

Capacitor with Dielectric:

When a dielectric is inserted:

• Capacitance increases: $C = \varepsilon_r C_0$
• Potential difference decreases: $V = \frac{V_0}{\varepsilon_r}$
• Stored energy changes: $U = \frac{1}{2}CV^2 = \frac{1}{2}\varepsilon_r C_0 \frac{V_0^2}{\varepsilon_r^2} = \frac{U_0}{\varepsilon_r}$

If the capacitor is disconnected (isolated), charge $Q$ stays constant and:

• $E$ decreases (since $V$ decreases)
• Energy decreases (work done by dielectric force)

If the capacitor is connected to a battery (constant voltage):

• $Q$ increases
• $E$ stays the same
• Energy increases (battery does work)

Gauss’s Law (Advanced):

$$\oint \vec{E} \cdot d\vec{A} = \frac{Q_{\text{enc}}}{\varepsilon_0}$$

The total electric flux through a closed surface equals the charge enclosed divided by $\varepsilon_0$.

This law is particularly useful for calculating electric fields of symmetric charge distributions.

⚡ NABTEB Quick Reference:

• $F = k\frac{q_1 q_2}{r^2}$, $k = 9 \times 10^9$ N·m²/C²
• $E = F/q = kQ/r^2$
• $V = kQ/r$
• $C = Q/V = \varepsilon_0\varepsilon_r A/d$
• $U = \frac{1}{2}CV^2 = \frac{1}{2}QV^2 = Q^2/2C$
• Series: $1/C_{\text{eq}} = \sum(1/C_i)$
• Parallel: $C_{\text{eq}} = \sum C_i$
• $E = -\frac{dV}{dr}$, or $E = V/d$ (uniform field)
• $\varepsilon_0 = 8.85 \times 10^{-12}$ F/m