Electric Field: Charges, Forces, Fields & Applications

TAKEAWAYS:

• Electric fields are invisible regions around charges where forces act on other charges.
• Field lines show direction (away from +, toward −) and density (stronger field = closer lines).
• Electric field strength (E) is force per unit charge: $E=\frac{F}{q}$ (measured in N/C).
• Coulomb’s law ($F=k\frac{q_1{q}_2}{r^2}$) explains force between two point charges.
• Electric dipoles (two equal/opposite charges) create non-uniform fields used in molecules and devices.
• Field mapping uses test charges to visualize fields around conductors and insulators.

1. Electric Charges and Forces

Electric charges are fundamental properties of matter. There are two types:

• Positive charge (+q): Gained by losing electrons (e.g., glass rubbed with silk).
• Negative charge (−q): Gained by gaining electrons (e.g., plastic rubbed with wool).

Key Properties:

• Like charges repel; unlike charges attract.
• Charge is quantized: smallest charge is $e=1.6\times 10^{-19}$ C (electron’s charge).
• Charge is conserved: total charge in an isolated system never changes.

Example 1.1: Charge Transfer When you comb your hair with a plastic comb, electrons move from your hair to the comb. Your hair becomes positively charged, and the comb becomes negatively charged. This causes your hair to stand up due to repulsion between like charges.

2. Coulomb’s Law: Force Between Charges

Coulomb’s law describes the electrostatic force between two point charges: $$F=k\frac{|q_1{q}_2|}{r^2}$$

• $F$: Force (N)
• $q_1,q_2$: Charges (C)
• $r$: Distance between charges (m)
• $k=9\times 10^9{\text{N m}}^2/{\text{C}}^2$ (Coulomb’s constant)

Direction:

• Force is attractive if charges are opposite.
• Force is repulsive if charges are like.

Example 2.1: Calculating Electrostatic Force Two charges $q_1=3\mu \text{C}$ and $q_2=-2\mu \text{C}$ are placed 0.5 m apart. Find the force between them. Solution: Convert charges to Coulombs: $q_1=3\times 10^{-6}\text{C}$, $q_2=-2\times 10^{-6}\text{C}$. Use Coulomb’s law: $$F=9\times 10^9\times \frac{|3\times 10^{-6}\times (-2)\times 10^{-6}|}{(0.5{)}^2}=216\text{N}$$ Since charges are opposite, the force is attractive.

3. Electric Field: Concept and Definition

An electric field is a region around a charged object where a force is exerted on other charges.

• Source charge (Q): Creates the field.
• Test charge (q₀): Used to detect the field (must be small so it doesn’t disturb the field).

Electric Field Strength (E): $$E=\frac{F}{q_0}$$

• Units: N/C (Newton per Coulomb).
• Direction: Same as force on a positive test charge.

Example 3.1: Field Due to a Point Charge A charge $Q=5\mu \text{C}$ creates a field. Find $E$ at 0.2 m. Solution: $$E=\frac{kQ}{r^2}=\frac{9\times 10^9\times 5\times 10^{-6}}{(0.2{)}^2}=1.125\times 10^6\text{N/C}$$ The field is radially outward (since $Q$ is positive).

4. Electric Field Lines

Field lines are imaginary lines that show:

1. Direction: Tangent to the line gives the field direction.
2. Strength: Closer lines = stronger field.

Rules for Drawing Field Lines:

• Start on positive charges, end on negative charges (or at infinity).
• Never cross or loop.
• Density is proportional to field strength.

Comparison Table: Field Lines for Different Charge Configurations

5. Electric Field Due to a System of Charges

For multiple charges, use the superposition principle: $${\vec{E}}_{\text{total}}={\vec{E}}_1+{\vec{E}}_2+{\vec{E}}_3+\dots$$ Each charge contributes independently to the total field.

Example 5.1: Field at the Midpoint of Two Charges Two charges $Q_1=4\mu \text{C}$ and $Q_2=-4\mu \text{C}$ are 6 cm apart. Find $E$ at the midpoint. Solution:

• Distance from midpoint to each charge: $r=3\text{cm}=0.03\text{m}$.
• Field due to $Q_1$: $E_1=\frac{9\times 10^9\times 4\times 10^{-6}}{(0.03{)}^2}=4\times 10^6\text{N/C}$ (rightward).
• Field due to $Q_2$: $E_2=\frac{9\times 10^9\times 4\times 10^{-6}}{(0.03{)}^2}=4\times 10^6\text{N/C}$ (leftward).
• Total field: $E=E_1-E_2=0\text{N/C}$ (fields cancel out).

6. Electric Dipole

An electric dipole consists of two equal and opposite charges ($+q$ and $-q$) separated by distance $2a$.

Key Properties:

• Dipole moment (p): $\vec{p}=q\times 2a$ (from −q to +q).
• Field along the axis: Stronger than perpendicular bisector.
• Field at large distances: Falls as $\frac{1}{r^3}$.

Example 6.1: Field on the Perpendicular Bisector A dipole has $q=2\mu \text{C}$, $2a=4\text{cm}$. Find $E$ at a point 3 cm from the center on the perpendicular bisector. Solution:

• Distance from each charge: $r=\sqrt{(0.02{)}^2+(0.03{)}^2}=0.036\text{m}$.
• Field due to $+q$: $E_{+}=\frac{9\times 10^9\times 2\times 10^{-6}}{(0.036{)}^2}=1.39\times 10^6\text{N/C}$ (angle θ = 37°).
• Field due to $-q$: $E_{-}=1.39\times 10^6\text{N/C}$ (angle θ = 37°).
• Vertical components cancel; horizontal components add: $E=2\times E_{+}\cos (37°)=1.74\times 10^6\text{N/C}$ (downward).

7. Applications of Electric Fields

1. Electrostatic Precipitators: Remove dust/ash from factory smoke using charged plates.
2. Photocopiers: Use static charges to transfer toner onto paper.
3. Capacitors: Store charge using electric fields between plates.
4. Biological Systems: Nerve impulses rely on ion movement in electric fields.

flowchart TD
    A["Electric Field Applications"] --> B["Industry: Precipitators"]
    A --> C["Technology: Photocopiers"]
    A --> D["Electronics: Capacitors"]
    A --> E["Biology: Nerve Signals"]

8. Solved NEB-Style Questions

Question 1: Short Answer Define electric field strength. How is it related to Coulomb’s law? Answer: Electric field strength ($E$) is the force per unit positive charge: $$E=\frac{F}{q_0}$$ From Coulomb’s law ($F=k\frac{qQ}{r^2}$), for a point charge $Q$: $$E=\frac{kQ}{r^2}$$

Question 2: Numerical Two charges $5\mu \text{C}$ and $-3\mu \text{C}$ are 10 cm apart. Find the electric field at a point 6 cm from the $5\mu \text{C}$ charge along the line joining them. Solution:

• Let $Q_1=5\mu \text{C}$, $Q_2=-3\mu \text{C}$, distance $d=10\text{cm}$.
• Point P is 6 cm from $Q_1$, so 4 cm from $Q_2$.
• Field due to $Q_1$: $E_1=\frac{9\times 10^9\times 5\times 10^{-6}}{(0.06{)}^2}=1.25\times 10^7\text{N/C}$ (rightward).
• Field due to $Q_2$: $E_2=\frac{9\times 10^9\times 3\times 10^{-6}}{(0.04{)}^2}=1.69\times 10^8\text{N/C}$ (leftward).
• Total field: $E=E_2-E_1=1.56\times 10^8\text{N/C}$ (leftward).

Question 3: Conceptual Why do electric field lines never intersect? Answer: If two field lines intersected, the field at that point would have two directions, which is impossible. Field lines represent a single direction of the field at every point.

Exam Tip

1. Memorize Coulomb’s law and field formulas—they appear in every numerical.
2. Draw field lines carefully for dipoles, single charges, and conductors.
3. Superposition is key: Break problems into contributions from each charge.
4. Units matter: Always convert $\mu \text{C}$ to C and cm to m.
5. Direction is critical: Use test charges to determine field direction.