Edexcel A Level (IAL) Physics-4.14 Electric Force between Two Charges- Study Notes- New Syllabus
Edexcel A Level (IAL) Physics -4.14 Electric Force between Two Charges- Study Notes- New syllabus
Edexcel A Level (IAL) Physics -4.14 Electric Force between Two Charges- Study Notes -Edexcel A level Physics – per latest Syllabus.
Key Concepts:
Force Between Two Point Charges (Coulomb’s Law)
The electric force between two stationary point charges acts along the line joining them. This force can be calculated using Coulomb’s law.
Coulomb’s Law
The magnitude of the force between two point charges is given by:
\( F = \dfrac{Q_1 Q_2}{4\pi \varepsilon r^2} \)
- \( F \) = electric force between the charges (N)
- \( Q_1, Q_2 \) = charges (C)
- \( r \) = separation between the charges (m)
- \( \varepsilon \) = permittivity of the medium (F m⁻¹)
In free space (or air):
\( \varepsilon = \varepsilon_0 = 8.85\times10^{-12}\ \mathrm{F\,m^{-1}} \)
Direction of the Force
- Like charges repel.
- Unlike charges attract.
- The force acts along the line joining the charges.
Important: The equation gives the magnitude of the force; direction must be stated separately.
Inverse Square Nature
- The force is proportional to the product \( Q_1Q_2 \).
- The force is inversely proportional to the square of the separation \( r^2 \).
- Doubling the distance reduces the force by a factor of four.
Relationship to Electric Fields
The force on a charge \( Q \) in an electric field \( E \) is:
\( F = EQ \)
Coulomb’s law explains the origin of electric fields produced by charges.
When to Use Coulomb’s Law
- Charges are stationary.
- Charges can be treated as point charges.
- Separation is much larger than the size of the charges.
Example (Easy)
Two charges of \( 2.0\times10^{-6}\ \mathrm{C} \) and \( 4.0\times10^{-6}\ \mathrm{C} \) are separated by \( 0.50\ \mathrm{m} \). Calculate the magnitude of the force between them.
▶️ Answer / Explanation
\( F = \dfrac{(2.0\times10^{-6})(4.0\times10^{-6})}{4\pi(8.85\times10^{-12})(0.50)^2} \)
\( F \approx 0.29\ \mathrm{N} \)
Example (Medium)
Two identical positive charges repel with a force of \( 0.10\ \mathrm{N} \) when separated by \( 0.30\ \mathrm{m} \). What is the force when the separation is increased to \( 0.60\ \mathrm{m} \)?
▶️ Answer / Explanation
- Force \( \propto \dfrac{1}{r^2} \)
- Distance doubled → force becomes one-quarter
\( F = \dfrac{0.10}{4} = 0.025\ \mathrm{N} \)
Example (Hard)
A positive charge and a negative charge experience an attractive force of \( 0.20\ \mathrm{N} \). Explain what happens to the force if the medium between them is replaced by a material with higher permittivity.
▶️ Answer / Explanation
- Force is inversely proportional to \( \varepsilon \).
- Higher permittivity increases the denominator.
- The force decreases.