7.2 Electrostatics, Coulomb's Law, Electric Fields & Capacitance

7.2 Electrostatics, Coulomb's Law, Electric Fields & Capacitance

Electrostatics investigates stationary electric charges, their interaction forces, electric fields, potential differences, and energy storage mechanisms in capacitors. This domain represents a major source of numerical and conceptual questions in the AMC Physics test.

---

1. Coulomb's Law & Dielectric Effects

Coulomb's Law quantifies the electrostatic force $F$ exerted between two point electric charges $q_1$ and $q_2$ separated by a distance $r$ in vacuum:

$F_{vac} = k \frac{|q_1 q_2|}{r^2} = \frac{1}{4\pi \varepsilon_0} \frac{|q_1 q_2|}{r^2}$

where:

• $k = \frac{1}{4\pi \varepsilon_0} \approx 8.99 \times 10^9 \text{ N}\cdot\text{m}^2/\text{C}^2$ is Coulomb's constant.
• $\varepsilon_0 \approx 8.85 \times 10^{-12} \text{ C}^2/(\text{N}\cdot\text{m}^2)$ is the permittivity of free space (vacuum).

Vector Form & Superposition

In vector form, the force exerted by charge $q_1$ on $q_2$ is:

$\vec{F}_{12} = \frac{1}{4\pi \varepsilon_0} \frac{q_1 q_2}{r^2} \hat{r}_{12}$

Like charges repel; opposite charges attract. By the Principle of Superposition, the net electrostatic force acting on a test charge due to a system of multiple point charges is the vector sum of individual Coulomb forces.

Effect of Dielectric Medium

When an insulating dielectric material of relative permittivity $\varepsilon_r$ (also called dielectric constant $K$) is placed between the charges, electric polarization reduces the net electrostatic force:

$F_{med} = \frac{F_{vac}}{\varepsilon_r} = \frac{1}{4\pi \varepsilon_0 \varepsilon_r} \frac{|q_1 q_2|}{r^2}$

Since relative permittivity $\varepsilon_r > 1$ for all material media, the presence of a dielectric medium always reduces electrostatic force.

---

2. Electric Field Intensity & Field Lines

Electric Field Intensity ((\vec{E}))

The electric field $\vec{E}$ at any point in space is defined as the electrostatic force experienced per unit positive test charge $q_0$ placed at that point:

$\vec{E} = \frac{\vec{F}}{q_0} \quad [\text{SI Units: N/C or V/m}]$

For a point charge $q$ generating the field at a radial distance $r$:

$E = \frac{1}{4\pi \varepsilon_0} \frac{|q|}{r^2}$

Field Lines Properties for AMC:

1. Emerge radially outward from positive charges and terminate at negative charges.
2. Tangent to a field line at any point gives the direction of $\vec{E}$.
3. Lines never intersect (if they did, $\vec{E}$ would have two directions at one point).
4. Field lines are perpendicular to the surface of a conductor in electrostatic equilibrium.
5. Electrostatic Shielding: The electric field inside a hollow or solid conductor in electrostatic equilibrium is strictly zero ($E = 0$).

---

3. Electric Potential & Potential Difference

Electric potential $V$ at a point is defined as the work done $W_{\infty \to P}$ per unit positive charge in bringing a test charge from infinity to that point without acceleration:

$V = \frac{W}{q_0} \quad [\text{SI Unit: Volt (V) = J/C}]$

For a point charge $q$:

$V = \frac{1}{4\pi \varepsilon_0} \frac{q}{r}$

Note that potential $V$ is a scalar quantity (unlike the vector electric field $\vec{E}$). Total potential due to multiple point charges is simply the algebraic sum of individual potentials.

Potential Gradient Relation

Electric field is related to potential variation via the negative potential gradient:

$E = -\frac{dV}{dr} \quad \text{or} \quad E = -\frac{\Delta V}{\Delta r}$

The negative sign indicates that the electric field vector $\vec{E}$ points in the direction of maximum decrease of electric potential.

Electron-Volt (eV) Unit

One electron-volt ($1 \text{ eV}$) is the energy gained or lost by an electron moving through a potential difference of $1 \text{ Volt}$:

$1 \text{ eV} = 1.602 \times 10^{-19} \text{ Joules}$

---

4. Gauss's Law & Electric Flux

Electric flux $\Phi_E$ through a surface of area $\vec{A}$ placed in a uniform electric field $\vec{E}$ is:

$\Phi_E = \vec{E} \cdot \vec{A} = E A \cos\theta \quad [\text{SI Unit: N}\cdot\text{m}^2/\text{C or V}\cdot\text{m}]$

Gauss's Law Statement

Gauss's Law states that the total electric flux $\Phi_E$ passing through any arbitrary closed Gaussian surface is equal to $\frac{1}{\varepsilon_0}$ times the total net charge $Q_{enclosed}$ enclosed within that surface:

$\Phi_E = \oint \vec{E} \cdot d\vec{A} = \frac{Q_{enclosed}}{\varepsilon_0}$

Key Applications of Gauss's Law:

1. Infinite Sheet of Charge: $E = \frac{\sigma}{2 \varepsilon_0}$ (where $\sigma$ is surface charge density).
2. Between Oppositely Charged Parallel Plates: $E = \frac{\sigma}{\varepsilon_0}$.
3. Inside a Conductive Shell: $E = 0$.

---

5. Capacitance & Energy Storage

A capacitor stores electric charge and electrostatic energy. Capacitance $C$ measures charge storage capability per unit potential difference:

$C = \frac{Q}{V} \quad [\text{SI Unit: Farad (F) = C/V}]$

Parallel Plate Capacitor

For two conducting plates of area $A$ separated by a distance $d$ in vacuum:

$C_0 = \frac{\varepsilon_0 A}{d}$

When a dielectric slab of relative permittivity $\varepsilon_r$ fills the region between plates:

$C = \frac{\varepsilon_r \varepsilon_0 A}{d} = \varepsilon_r C_0$

Inserting a dielectric increases capacitance by a factor of $\varepsilon_r$.

Combinations of Capacitors

Stored Electrostatic Energy

The work done in charging a capacitor is stored as electrostatic potential energy $U$ in the electric field between plates:

$U = \frac{1}{2} C V^2 = \frac{1}{2} Q V = \frac{Q^2}{2C}$

Energy density $u$ (energy per unit volume) in an electric field $E$ is given by $u = \frac{1}{2} \varepsilon_0 E^2$.

---

6. Worked Numerical Examples for AMC Candidates

Example 1: Electrostatic Force Between Point Charges

Problem: Two point charges $q_1 = +2 ,\mu\text{C}$ and $q_2 = -4 ,\mu\text{C}$ are placed in vacuum separated by $0.3 \text{ m}$. Calculate the magnitude of attraction force.

Solution:

1. Convert charges: $q_1 = 2 \times 10^{-6} \text{ C}$, $q_2 = 4 \times 10^{-6} \text{ C}$.
2. Apply Coulomb's Law: $F = (8.99 \times 10^9) \frac{(2 \times 10^{-6})(4 \times 10^{-6})}{(0.3)^2} = (8.99 \times 10^9) \frac{8 \times 10^{-12}}{0.09}$ $F = \frac{0.07192}{0.09} \approx 0.799 \text{ N}$

Example 2: Equivalent Capacitance and Energy

Problem: Two capacitors of $6 ,\mu\text{F}$ and $12 ,\mu\text{F}$ are connected in series across a $100 \text{ V}$ DC supply. Find the equivalent capacitance and total stored energy.

Solution:

1. Series equivalent capacitance: $\frac{1}{C_{eq}} = \frac{1}{6} + \frac{1}{12} = \frac{2 + 1}{12} = \frac{3}{12} = \frac{1}{4} \implies C_{eq} = 4 \,\mu\text{F} = 4 \times 10^{-6} \text{ F}$
2. Stored energy: $U = \frac{1}{2} C_{eq} V^2 = \frac{1}{2} (4 \times 10^{-6} \text{ F}) (100 \text{ V})^2 = (2 \times 10^{-6}) \times 10000 = 0.02 \text{ J}$