Electric field of a point charge

1. Overview

A point charge is a theoretical model of a charged object where all its electric charge is concentrated at a single, infinitesimal point in space. In reality, this model is applied to any charged body whose physical dimensions are negligible compared to the distance at which the field is being measured.

The electric field of a point charge is radial and obeys an inverse square law. This means the electric field strength ($E$) decreases rapidly as the distance ($r$) from the charge increases. The field is modeled as extending infinitely into space, though its strength becomes negligible at large distances. For a single point charge in free space (a vacuum), the field strength is determined solely by the magnitude of the charge and the distance from it, mediated by the permittivity of free space ($\varepsilon_0$).

---

Key Definitions

• Electric Field: A region of space where a stationary charge experiences an electric force.
• Electric Field Strength ($E$): The force per unit positive charge acting at a point in an electric field.
  • It is a vector quantity; its direction is the direction of the force on a positive test charge.
  • Units: Newtons per Coulomb ($\text{N C}^{-1}$) or Volts per metre ($\text{V m}^{-1}$).
• Point Charge: A charged body where the dimensions are negligible compared to the separation between the body and the point of observation.
• Permittivity of Free Space ($\varepsilon_0$): A physical constant that describes the ability of a vacuum to permit electric field lines.
  • Value: $\approx 8.85 \times 10^{-12} \text{ F m}^{-1}$.
• Radial Field: A field in which the field lines are straight, non-parallel, and either converge toward or diverge away from a single point.
• Inverse Square Law: A physical law stating that a specified physical quantity is inversely proportional to the square of the distance from the source of that quantity.

---

Content

3.1 The Point Charge Model and Spherical Conductors

In Cambridge 9702, the point charge model is frequently applied to uniformly charged spheres.

• Outside the sphere ($r \geq R$): A uniformly charged conducting or non-conducting sphere creates an electric field identical to that of a point charge located at the sphere's geometric center.
• Inside the sphere ($r < R$): For a conducting sphere in electrostatic equilibrium, the electric field strength is zero because all excess charge resides on the outer surface and the internal forces cancel out.

3.2 Derivation of the Electric Field Strength Formula

The electric field strength $E$ is derived from the definition of the field and Coulomb’s Law.

1. Definition of $E$: The field strength is the force $F$ per unit test charge $+q$. $$\mathbf{E = \frac{F}{q}}$$

2. Coulomb’s Law: The force between a source charge $Q$ and a test charge $q$ separated by distance $r$ is: $$\mathbf{F = \frac{Qq}{4\pi\varepsilon_0 r^2}}$$

3. Substitution: Substitute the force expression into the definition of $E$: $$E = \frac{\left( \frac{Qq}{4\pi\varepsilon_0 r^2} \right)}{q}$$

4. Result: The test charge $q$ cancels out, leaving the field strength produced by the source charge $Q$: $$\mathbf{E = \frac{Q}{4\pi\varepsilon_0 r^2}}$$

Directional Rules:

• If $Q$ is positive ($+Q$): The field lines point radially outwards.
• If $Q$ is negative ($-Q$): The field lines point radially inwards.

3.3 The Inverse Square Law ($E \propto 1/r^2$)

The relationship $E \propto \frac{1}{r^2}$ is a defining characteristic of radial fields.

• If the distance $r$ is doubled, the field strength $E$ decreases by a factor of 4 ($2^2$).
• If the distance $r$ is tripled, the field strength $E$ decreases by a factor of 9 ($3^2$).
• If the distance $r$ is halved, the field strength $E$ increases by a factor of 4.

3.4 Graphical Representations

Students must be able to recognize and sketch two primary graphs for a point charge:

1. $E$ against $r$:
   • The graph is a curve (hyperbola-like) in the first quadrant for a positive charge.
   • $E \to \infty$ as $r \to 0$ (though physically, the point charge model breaks down at $r=0$).
   • $E \to 0$ as $r \to \infty$ (the x-axis is an asymptote).
2. $E$ against $1/r^2$:
   • This produces a straight line through the origin.
   • The gradient of this line is equal to $\frac{Q}{4\pi\varepsilon_0}$.

3.5 Field Line Geometry

Field lines (lines of force) provide a visual map of the electric field:

• Density: The closer the lines are to each other, the stronger the electric field. In a radial field, line density is highest at the surface of the charge and decreases as $1/r^2$ as they spread over the surface of an imaginary sphere of area $4\pi r^2$.
• Intersection: Field lines never cross. If they did, a single point would have two different field directions, which is physically impossible.
• Surface Interaction: Field lines always meet the surface of a conductor at right angles ($90^\circ$).

3.6 Superposition of Electric Fields

When multiple point charges are present, the net electric field strength at any point is the vector sum of the individual fields created by each charge.

$$\mathbf{E_{net} = \vec{E_1} + \vec{E_2} + \dots + \vec{E_n}}$$

Solving Superposition Problems:

1. Calculate the magnitude of $E$ from each charge using $E = \frac{Q}{4\pi\varepsilon_0 r^2}$.
2. Determine the direction of each field (away from positive, toward negative).
3. If the fields are collinear (on the same line): Add them as signed scalars (e.g., right is positive, left is negative).
4. If the fields are at an angle: Resolve them into horizontal ($x$) and vertical ($y$) components, sum the components, and use Pythagoras' theorem: $E_{net} = \sqrt{E_x^2 + E_y^2}$.

---

3.7 Worked Example 1: Single Point Charge

Question: A point charge of $-4.5 \text{ }\mu\text{C}$ is situated in a vacuum. Calculate the electric field strength at a point $P$ located $12 \text{ cm}$ from the charge. State the direction.

Step 1: Convert units to SI $Q = -4.5 \times 10^{-6} \text{ C}$ $r = 0.12 \text{ m}$ $\varepsilon_0 = 8.85 \times 10^{-12} \text{ F m}^{-1}$

Step 2: State the equation $$E = \frac{Q}{4\pi\varepsilon_0 r^2}$$

Step 3: Substitute values $$E = \frac{4.5 \times 10^{-6}}{4 \times \pi \times 8.85 \times 10^{-12} \times (0.12)^2}$$ (Note: We use the magnitude of $Q$ for the calculation and determine direction separately.)

Step 4: Calculate $$E = \frac{4.5 \times 10^{-6}}{1.602 \times 10^{-12}}$$ $$E = 2.81 \times 10^6 \text{ N C}^{-1}$$

Step 5: Determine direction Since the charge is negative, the field acts radially inwards (towards the charge).

---

3.8 Worked Example 2: Point of Zero Field Strength (Neutral Point)

Question: Two point charges $A$ and $B$ are separated by a distance of $15.0 \text{ cm}$ in a vacuum. Charge $A$ is $+2.0 \text{ nC}$ and charge $B$ is $+8.0 \text{ nC}$. Find the distance from charge $A$ where the resultant electric field strength is zero.

Step 1: Conceptualize For the net field to be zero, the fields from $A$ and $B$ must be equal in magnitude and opposite in direction. Since both are positive, this point must lie between them. Let $x$ be the distance from $A$. The distance from $B$ is $(0.15 - x)$.

Step 2: Set $E_A = E_B$ $$\frac{Q_A}{4\pi\varepsilon_0 x^2} = \frac{Q_B}{4\pi\varepsilon_0 (0.15 - x)^2}$$

Step 3: Simplify Cancel $4\pi\varepsilon_0$ from both sides: $$\frac{2.0 \times 10^{-9}}{x^2} = \frac{8.0 \times 10^{-9}}{(0.15 - x)^2}$$ $$\frac{2}{x^2} = \frac{8}{(0.15 - x)^2}$$

Step 4: Take the square root of both sides $$\frac{\sqrt{2}}{x} = \frac{\sqrt{8}}{0.15 - x}$$ $$\frac{1.414}{x} = \frac{2.828}{0.15 - x}$$

Step 5: Solve for $x$ $$1.414(0.15 - x) = 2.828x$$ $$0.2121 - 1.414x = 2.828x$$ $$0.2121 = 4.242x$$ $$x = 0.050 \text{ m} = 5.0 \text{ cm}$$

Answer: The field is zero at $5.0 \text{ cm}$ from charge $A$.

---

3.9 Worked Example 3: Field from a Charged Sphere

Question: A metal sphere of radius $R = 3.0 \text{ cm}$ carries a charge of $+6.0 \text{ nC}$. Calculate the electric field strength at: a) $r = 2.0 \text{ cm}$ from the center. b) $r = 5.0 \text{ cm}$ from the center.

Part (a) Solution: The point $r = 2.0 \text{ cm}$ is inside the conducting sphere ($r < R$). In a conductor, the electric field strength is zero. $E = 0 \text{ N C}^{-1}$.

Part (b) Solution: The point $r = 5.0 \text{ cm}$ is outside the sphere ($r > R$). We treat the sphere as a point charge at its center. $Q = 6.0 \times 10^{-9} \text{ C}$ $r = 0.05 \text{ m}$ $$E = \frac{6.0 \times 10^{-9}}{4\pi(8.85 \times 10^{-12})(0.05)^2}$$ $$E = \frac{6.0 \times 10^{-9}}{2.78 \times 10^{-13}}$$ $E = 2.16 \times 10^4 \text{ N C}^{-1}$.

---

Key Equations

Equation	Description	Data Sheet?
$E = \frac{Q}{4\pi\varepsilon_0 r^2}$	Electric field strength of a point charge.	Yes
$F = qE$	Force on a charge $q$ in an electric field $E$.	Yes
$E = \frac{V}{d}$	Field strength in a uniform field (for comparison).	Yes
$k = \frac{1}{4\pi\varepsilon_0}$	Coulomb constant ($\approx 8.99 \times 10^9 \text{ N m}^2 \text{ C}^{-2}$).	No

---

Common Mistakes to Avoid

• ❌ The "Square" Error: Forgetting to square the distance $r$ in the denominator.
  • ✓ Right: Always write $r^2$ first before plugging in numbers.
• ❌ Radius vs. Distance: Using the distance from the surface of a sphere instead of the distance from the center.
  • ✓ Right: $r$ is always the distance from the center of the point charge or sphere. If a question says "5 cm from the surface of a 2 cm radius sphere," $r = 7 \text{ cm}$.
• ❌ Unit Neglect: Leaving charge in $\mu\text{C}$ or distance in $\text{cm}$.
  • ✓ Right: Convert everything to Coulombs ($\text{C}$) and Metres ($\text{m}$) immediately.
    • $1 \text{ }\mu\text{C} = 10^{-6} \text{ C}$
    • $1 \text{ nC} = 10^{-9} \text{ C}$
    • $1 \text{ pC} = 10^{-12} \text{ C}$
• ❌ Scalar Addition of Vectors: Adding the magnitudes of fields from two charges when they are not pointing in the same direction.
  • ✓ Right: Draw a vector diagram. If fields point in opposite directions, subtract them. If at an angle, use components.
• ❌ Confusing $Q$ and $q$: Using the test charge magnitude to calculate the field strength.
  • ✓ Right: $Q$ is the charge creating the field. $q$ is the charge experiencing the force. The field $E$ does not depend on $q$.

---

Exam Tips

1. Field Inside a Conductor: If a question asks for the field inside a metal sphere or any conductor, the answer is almost always zero. Don't waste time calculating.
2. Graphing $E$ vs $r$: When sketching, ensure the curve never touches the x or y axes. If the graph is for a sphere of radius $R$, draw a solid line at $E=0$ from $r=0$ to $r=R$, then a vertical dotted line up to the maximum value at $r=R$, then the $1/r^2$ curve.
3. Comparison to Gravity: Use your knowledge of Gravitational Fields ($g = \frac{GM}{r^2}$) to check your work. The math is identical, but remember that electric fields can be attractive or repulsive, whereas gravity is only attractive.
4. Significant Figures: Cambridge 9702 usually requires answers to 2 or 3 significant figures. Look at the data provided (e.g., if $r = 5.0 \text{ cm}$, use 2 s.f.).
5. The $4\pi$ Factor: Remember that $4\pi r^2$ is the surface area of a sphere. This is why $4\pi$ appears in the radial field formula but not in the uniform field formula ($E = V/d$).
6. Ratio Method: For multiple-choice questions, use ratios. If $r$ triples, $E$ becomes $1/9$. You don't need to calculate the full value of $1/4\pi\varepsilon_0$ every time.