Edexcel A Level (IAL) Physics-4.15 Electric Field due to a Point Charge- Study Notes- New Syllabus
Edexcel A Level (IAL) Physics -4.15 Electric Field due to a Point Charge- Study Notes- New syllabus
Edexcel A Level (IAL) Physics -4.15 Electric Field due to a Point Charge- Study Notes -Edexcel A level Physics – per latest Syllabus.
Key Concepts:
- be able to use the equation \(E=\dfrac{Q}{4\pi\varepsilon_0 r^2}\) for the electric field due to a point charge
Electric Field Due to a Point Charge
The electric field produced by a single stationary point charge can be calculated using a form of Coulomb’s law. This describes how the strength of the field varies with distance from the charge.
Equation for Electric Field Due to a Point Charge
The magnitude of the electric field at distance \( r \) from a point charge \( Q \) is:
\( E = \dfrac{Q}{4\pi \varepsilon r^2} \)
- \( E \) = electric field strength (N C⁻¹)
- \( Q \) = charge producing the field (C)
- \( r \) = distance from the charge (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 Electric Field
- For a positive charge, the electric field points radially away from the charge.
- For a negative charge, the electric field points radially towards the charge.
- The equation gives the magnitude of the field; direction must be stated separately.
Inverse Square Dependence
- Electric field strength is proportional to the charge \( Q \).
- Electric field strength is inversely proportional to \( r^2 \).
- Doubling the distance reduces the field strength by a factor of four.
Relationship to Force on a Charge
The force on a charge \( q \) placed in this electric field is:
\( F = Eq \)
This shows that the field exists independently of the test charge.
When This Equation Applies
- The charge producing the field is stationary.
- The charge can be treated as a point charge.
- The distance from the charge is much larger than the size of the charge.
Example (Easy)
Calculate the electric field strength \( 0.40\ \mathrm{m} \) from a point charge of \( 2.0\times10^{-6}\ \mathrm{C} \) in air.
▶️ Answer / Explanation
\( E = \dfrac{2.0\times10^{-6}}{4\pi(8.85\times10^{-12})(0.40)^2} \)
\( E \approx 1.12\times10^{5}\ \mathrm{N\,C^{-1}} \)
Example (Medium)
The electric field strength at a point is \( 2.0\times10^{4}\ \mathrm{N\,C^{-1}} \). Find the distance from a point charge of \( 1.0\times10^{-6}\ \mathrm{C} \).
▶️ Answer / Explanation
Rearrange the equation:
\( r^2 = \dfrac{Q}{4\pi\varepsilon E} \)
\( r = \sqrt{\dfrac{1.0\times10^{-6}}{4\pi(8.85\times10^{-12})(2.0\times10^{4})}} \)
\( r \approx 0.67\ \mathrm{m} \)
Example (Hard)
Explain how the electric field strength around a point charge changes if the charge is doubled and the distance is halved.
▶️ Answer / Explanation
- Doubling \( Q \) doubles the field strength.
- Halving \( r \) increases the field strength by a factor of four.
- Total increase factor = \( 2 \times 4 = 8 \).
- The field strength increases eightfold.