Explain the concept, theories and principles of electrostatics (electric field, electric potential and capacitance)

2.1 Definition

The electric field $\vec{E}$ at a point is the electrostatic force per unit positive test charge placed at that point:

$\vec{E}= \frac{\vec{F}}{q}.$

The unit of $E$ is newton per coulomb ($\text{N C}^{-1}$) or volt per metre ($\text{V m}^{-1}$).

2.2 Field due to a point charge

For a point charge $Q$ at the origin, the magnitude of the field at a distance $r$ is

$E = \frac{kQ}{r^2}.$

The direction is radially outward from $Q$ if $Q$ is positive and radially inward if $Q$ is negative.

2.3 Superposition principle

When several charges are present, the total field is the vector sum of the fields due to each charge:

$\vec{E}_{\text{total}} = \sum_i \vec{E}_i.$

2.4 Electric field lines

• Field lines start on positive charges and end on negative charges.
• The tangent to a line at any point gives the direction of $\vec{E}$.
• Where lines are close together the field is strong; where they are far apart the field is weak.

Worked example 2

Two point charges of $+2\,\mu\text{C}$ each are placed 20 cm apart. What is the electric field at the midpoint?

Both charges have the same magnitude and are equidistant from the midpoint, so their fields at that point are equal in magnitude but opposite in direction. Hence they cancel:

$E_{\text{midpoint}} = 0.$

2.5 Gauss’s law (qualitative)

The total electric flux through a closed surface equals the net charge inside divided by $\varepsilon_0$:

$\Phi_E = \frac{Q_{\text{inside}}}{\varepsilon_0}.$

Gauss’s law is especially useful for highly symmetric charge distributions (spherical, cylindrical, planar).

3.1 Definition

The electric potential $V$ at a point is the work done per unit positive charge in bringing a test charge from infinity to that point (assuming zero potential at infinity).

$V = \frac{W}{q}.$

The unit of potential is the volt ($\text{V}$), where $1\;\text{V}=1\;\text{J C}^{-1}$.

3.2 Potential due to a point charge

For a point charge $Q$ at the origin,

$V = \frac{kQ}{r}.$

3.3 Superposition for potentials

Potential is a scalar, so the total potential at a point is the algebraic sum of the potentials due to each charge:

$V_{\text{total}} = \sum_i V_i.$

3.4 Relationship between field and potential

For a uniform field,

$V_a - V_b = -\int_a^b \vec{E}\cdot d\vec{r}.$

In one dimension, $E = -\dfrac{dV}{dr}$. Thus the electric field points in the direction of decreasing potential.

3.5 Equipotential surfaces

An equipotential surface is a surface on which the potential is the same everywhere. Key properties:

• No work is required to move a charge along an equipotential.
• Electric field lines are always perpendicular to equipotential surfaces.

3.6 Electric potential energy

The electrostatic potential energy $U$ of a charge $q$ at a point where the potential is $V$ is

$U = qV.$

For a system of point charges, the total potential energy is the work required to assemble the configuration from infinity.

Worked example 3

Two point charges $+20\,\mu\text{C}$ and $-20\,\mu\text{C}$ are 20 cm apart. Find the potential at the midpoint.

At the midpoint $r = 10\;\text{cm}=0.10\;\text{m}$ from each charge:

$V = \frac{k(+20\times10^{-6})}{0.10} + \frac{k(-20\times10^{-6})}{0.10} = \frac{9\times10^9\times20\times10^{-6}}{0.10} - \frac{9\times10^9\times20\times10^{-6}}{0.10}=0\;\text{V}.$

4.1 Definition

A capacitor consists of two conducting plates separated by an insulator (dielectric). The capacitance $C$ is defined as the ratio of the magnitude of charge $Q$ on either plate to the potential difference $V$ between them:

$C = \frac{Q}{V}.$

The unit of capacitance is the farad ($\text{F}$); $1\;\text{F}=1\;\text{C V}^{-1}$. Typical values are microfarad ($\mu\text{F}=10^{-6}\,\text{F}$) or picofarad ($\text{pF}=10^{-12}\,\text{F}$).

4.2 Parallel‑plate capacitor

For a parallel‑plate capacitor with plate area $A$ and plate separation $d$ (with air or vacuum between the plates),

$C = \frac{\varepsilon_0 A}{d}.$

If a dielectric of relative permittivity $\varepsilon_r$ completely fills the gap, the capacitance becomes

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

where $C_0$ is the capacitance without the dielectric.

4.3 Factors affecting capacitance

• Plate area $A$ – larger area ⇒ larger $C$ (directly proportional).
• Plate separation $d$ – smaller separation ⇒ larger $C$ (inversely proportional).
• Dielectric constant $\varepsilon_r$ – larger $\varepsilon_r$ ⇒ larger $C$ (directly proportional).

4.4 Energy stored in a capacitor

The work done to charge a capacitor is stored as electric potential energy. For a capacitor charged to voltage $V$:

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

The energy density (energy per unit volume) in the field between the plates is

$\eta = \frac{1}{2}\varepsilon_0 E^2.$

4.5 Combination of capacitors

• Series: $\displaystyle \frac{1}{C_{\text{eq}}}= \frac{1}{C_1}+ \frac{1}{C_2}+ \cdots$
• Parallel: $C_{\text{eq}} = C_1 + C_2 + \cdots$

Worked example 4

Two capacitors $C_1=6.0\,\mu\text{F}$ and $C_2=3.0\,\mu\text{F}$ are connected in series across an 18 V battery. Find (a) the equivalent capacitance, (b) the charge on each capacitor.

(a) For series:

$C_{\text{eq}} = \frac{C_1C_2}{C_1+C_2}= \frac{6\times3}{6+3}=2.0\,\mu\text{F}.$

(b) The total charge is

$Q = C_{\text{eq}}V = 2.0\times10^{-6}\times18 = 3.6\times10^{-5}\,\text{C}.$

In a series combination the same charge resides on each capacitor, so each has $Q = 3.6\times10^{-5}\,\text{C}$.

4.6 Charging and discharging (RC circuits)

When a capacitor is connected to a resistor $R$ and a battery, the charge grows as

$Q(t) = CV\!\left(1-e^{-t/RC}\right).$

The product $RC$ is the time constant $\tau$; it measures how fast the capacitor charges. After a time $t=\tau$, the capacitor reaches about $63\%$ of its final charge.

During discharge through a resistor,

$Q(t) = Q_0e^{-t/RC},\qquad I(t)=\frac{Q_0}{RC}e^{-t/RC}.$

These equations constitute the core theories and principles of electrostatics that enable us to analyse and solve a wide range of practical problems.

In Tanzania, the concept of capacitance is applied in capacitive soil‑moisture sensors used by small‑scale farmers. The sensor consists of two metal plates embedded in the soil; the soil’s water content acts as a dielectric with a relative permittivity around 80, much higher than dry soil (≈ 4). By measuring the change in capacitance, a farmer can estimate soil moisture and decide when to irrigate, thereby saving water and improving crop yields. This inexpensive technique is already being promoted in some irrigation projects across the country.