Electric field

Electric field $\mathbf{E}$ is analogous to gravitational field $\mathbf{g}$. It is a region around a charge where another charge experiences a force. The electric field intensity at a point is defined as:

$\mathbf{E} = \frac{\mathbf{F}}{q}$

Coulomb's law

The electrostatic force between two point charges is given by:

$F = k \frac{Q_1 Q_2}{r^2} \quad \text{(9.1)}$

where $k = \frac{1}{4 \pi \varepsilon_0}$ and $\varepsilon_0 = 8.854 \times 10^{-12} \; \text{C}^2 \text{N}^{-1} \text{m}^{-2}$

So:

$F = \frac{1}{4 \pi \varepsilon_0} \frac{Q_1 Q_2}{r^2} \quad \text{(9.2)}$

Relative permittivity:

$\varepsilon_r = \frac{\varepsilon}{\varepsilon_0}$

Vector form:

$\mathbf{F}_{21} = k \frac{Q_1 Q_2}{r^3} \mathbf{r} \quad \text{(9.3)}$

Superposition principle

$\mathbf{F} = \sum_{i=1}^{n} \mathbf{F}_i$

$F = \sqrt{\left(\sum F_x\right)^2 + \left(\sum F_y\right)^2}$

Example 1

$Q = 1 \times 10^{-7} \text{ C}, \; e = 1.6 \times 10^{-19} \text{ C}$

$n = \frac{Q}{e} = \frac{1 \times 10^{-7}}{1.6 \times 10^{-19}} = 6.25 \times 10^{11} \text{ electrons}$

Example 2

$r = 5.3 \times 10^{-11} \text{ m}, \; Q_1 = Q_2 = 1.6 \times 10^{-19} \text{ C}$

$F = \frac{9.0 \times 10^9 (1.6 \times 10^{-19})^2}{(5.3 \times 10^{-11})^2} \approx 8.2 \times 10^{-8} \text{ N}$

Example 3

Three charges: $Q_1 = 1.0 \text{ nC}, Q_2 = 3.0 \text{ nC}, Q_3 = 5.0 \text{ nC}$

Distances: $r_{31} = 0.02 \text{ m}, r_{32} = 0.04 \text{ m}$

$F_{31} = \frac{9 \times 10^9 \cdot 1 \cdot 5 \times 10^{-18}}{(0.02)^2} = 1.13 \times 10^{-4} \text{ N}$

$F_{32} = \frac{9 \times 10^9 \cdot 3 \cdot 5 \times 10^{-18}}{(0.04)^2} = 8.44 \times 10^{-5} \text{ N}$

$\sum F = F_{31} + F_{32} = 1.97 \times 10^{-4} \text{ N (left)}$

Example 4

Triangle side = 0.2 m. Charges: 2μC, 3μC, 4μC

$F = \frac{9 \times 10^9 \cdot 4 \cdot 3 \times 10^{-12}}{(0.2)^2} = 2.7 \text{ N}$

Net vertical force: $-F \sin 60^\circ = -1.56 \text{ N}$

Net horizontal force: $F \cos 60^\circ + F_{2\mu C} = 3.6 \text{ N}$

Electric field of a point charge

$E = \frac{F}{q} \quad \text{(9.4)}$

Using Coulomb's law:

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

Then,

$E = \frac{1}{4 \pi \varepsilon_0} \frac{q_o}{r^2} \quad \text{(9.5)}$

Electric field lines

Originate from positive and terminate on negative charges. They never cross each other and are perpendicular to conductors' surfaces.

Example

Two charges $2 \mu \text{C}$ at 20 cm apart:

At midpoint, $E_1 = E_2$, but in opposite directions ⇒ $\sum E = 0$

Example

$Q = 3.3 \text{ nC}, \varepsilon_r = 5, r = 0.1 \text{ m}$

$\varepsilon = \varepsilon_0 \varepsilon_r = 5 \cdot 8.854 \times 10^{-12}$

$E = \frac{1}{4 \pi \varepsilon} \frac{Q}{r^2} = \frac{9 \times 10^9}{5} \cdot \frac{3.3 \times 10^{-9}}{0.01} \approx 594 \; \text{N/C}$

Electric field due to continuous charge distribution

In practice, we deal with charge distribution along a line, over a surface, or within a volume. The linear, surface, and volume charge densities are:

$\lambda = \frac{dQ}{dl}, \quad \sigma = \frac{dQ}{dA}, \quad \rho = \frac{dQ}{dV}$

These represent the charge per unit length, area, and volume, respectively. The electric field due to a continuous distribution is the sum (integral) of infinitesimal fields $dE$ from elements $dQ$.

Vector form: The infinitesimal electric field due to an element $dQ$ at position $\vec{r}'$ is:

$d\vec{E} = \frac{1}{4\pi\varepsilon_0} \frac{dQ (\vec{r} - \vec{r}')}{|\vec{r} - \vec{r}'|^3}$

and the total electric field is:

$\vec{E} = \frac{1}{4\pi\varepsilon_0} \int \frac{(\vec{r} - \vec{r}')}{|\vec{r} - \vec{r}'|^3} \, dq$

a. Electric field due to a line of charge

$dE = k \frac{dQ}{r^2}, \quad dE_x = dE \cos \theta = k \frac{dQ}{r^2} \frac{x}{r}$

$dQ = \lambda dy$

Total field:

$E_p = \int_{-a}^{b} dE_x = k\lambda \int_{-a}^{b} \frac{x \, dy}{(x^2 + y^2)^{3/2}}$

$= \frac{k\lambda x}{\sqrt{x^2 + y^2}} \Big|_{-a}^{b} = \frac{k\lambda x}{\sqrt{x^2 + b^2}} + \frac{k\lambda x}{\sqrt{x^2 + a^2}}$

As $a, b \to \infty$:

$E_p = \frac{2k\lambda}{x} = \frac{\lambda}{2\pi \varepsilon_0 x}$

Note: This result demonstrates the inverse dependence on distance from an infinite line of charge, unlike the inverse-square law for point charges.

Electric flux and Gauss's law

$\Phi = \vec{E} \cdot \vec{A} = EA \cos \theta$

For curved surfaces or non-uniform fields:

$\Phi = \int E \cos \theta \, dA = \int \vec{E} \cdot d\vec{A}$

Example 8

$\Phi = E A \cos 30^\circ = (2.0 \times 10^3)(\pi \cdot 0.1^2) \cos 30^\circ \approx 54 \, \text{Nm}^2\text{/C}$

Gauss's law

$\oint \vec{E} \cdot d\vec{A} = \frac{Q}{\varepsilon_0}$

Steps:

1. Choose symmetric Gaussian surface
2. Ensure $\vec{E} \parallel d\vec{A}$
3. Integrate over the entire surface

Advanced symmetries: Gauss's law is powerful when applied to systems with high symmetry:

• Plane symmetry → Infinite plane sheet
• Cylindrical symmetry → Line charge, coaxial cables
• Spherical symmetry → Spherical shells, point charges

Limitation: Gauss's law is always true, but only useful for calculating $\vec{E}$ when symmetry simplifies the surface integral.

b. Electric field due to a spherical charge

For a solid sphere (radius R, charge density $\rho$):

$Q = \frac{4}{3}\pi R^3 \rho$

Outside ($r > R$):

$E = \frac{Q}{4\pi\varepsilon_0 r^2}$

Inside ($r < R$):

$Q' = Q \left( \frac{r^3}{R^3} \right), \quad E = \frac{Qr}{4\pi\varepsilon_0 R^3}$

Graphical insight: Electric field increases linearly inside a uniformly charged sphere, reaches maximum at the surface, then decreases as $1/r^2$ outside.

c. Solid conducting sphere

Outside ($r > R$):

$E = \frac{Q}{4\pi\varepsilon_0 r^2}$

Inside ($r < R$):

$E = 0$

Reason: In electrostatics, charges on conductors reside on the surface. The field inside a conductor in equilibrium is zero.

d. Hollow conducting sphere

Outside ($r > R_2$):

$E = \frac{Q}{4\pi\varepsilon_0 r^2}$

Inside hollow ($r < R_1$):

$E = 0$

Between shells ($R_1 < r < R_2$):

$E = \frac{Q(r^3 - R_1^3)}{4\pi\varepsilon_0 r^2 (R_2^3 - R_1^3)}$

Note: Inside a cavity of a conductor, the electric field is zero if no charge is present inside.

f. Electric field due to plane charge

For an infinite plane:

$E = \frac{\sigma}{2\varepsilon_0}$

Two sheets with $\sigma_1, \sigma_2$:

Region I: $E = -\frac{1}{2\varepsilon_0}(\sigma_1 + \sigma_2)$

Region II: $E = -\frac{1}{2\varepsilon_0}(\sigma_1 - \sigma_2)$

Region III: $E = \frac{1}{2\varepsilon_0}(\sigma_1 + \sigma_2)$

Parallel plate capacitors mimic plane charge distribution behavior. Between the plates, the field is:

$E = \frac{\sigma}{\varepsilon_0} \quad \text{(if equal and opposite)}$

This uniform field is essential in applications like CRTs, capacitors, and sensors.

Additional concept: Superposition of electric fields

The principle of superposition allows for calculating the net electric field due to multiple sources:

$\vec{E}_{\text{net}} = \vec{E}_1 + \vec{E}_2 + \vec{E}_3 + \cdots$

This is crucial when dealing with complex configurations of charges.

Application: Charged cylindrical shell (Advanced Gauss's law)

Consider a long cylindrical shell of radius $R$ carrying surface charge density $\sigma$. Using a cylindrical Gaussian surface:

• $r > R$: $E = \frac{\sigma R}{\varepsilon_0 r}$
• $r < R$: $E = 0$