CIE AS/A Level Physics 18.4 Electric field of a point charge Study Notes- 2025-2027 Syllabus
CIE AS/A Level Physics 18.4 Electric field of a point charge Study Notes – New Syllabus
CIE AS/A Level Physics 18.4 Electric field of a point charge Study Notes at IITian Academy focus on specific topic and type of questions asked in actual exam. Study Notes focus on AS/A Level Physics latest syllabus with Candidates should be able to:
- recall and use E = Q / (4πε0 r 2) for the electric field strength due to a point charge in free space
Electric Field Strength Due to a Point Charge
For a single point charge in free space, the electric field strength at a distance \( \mathrm{r} \) from the charge is given by:
\( \mathrm{E = \dfrac{Q}{4\pi\epsilon_0 r^2}} \)
- \( \mathrm{E} \) = electric field strength (N C⁻¹ or V m⁻¹)
- \( \mathrm{Q} \) = point charge (C)
- \( \mathrm{r} \) = distance from the charge (m)
- \( \mathrm{\epsilon_0 = 8.85\times 10^{-12}\ F\,m^{-1}} \)
- \( \mathrm{\dfrac{1}{4\pi\epsilon_0} = 9.0\times 10^9\ N\,m^2\,C^{-2}} \)
Key Points:
- The field decreases with the square of distance → inverse square law.
- Field direction:
- away from positive charges
- toward negative charges
- The formula is valid only for points outside the charge distribution.
Example
Calculate the electric field strength 0.40 m from a \( \mathrm{+3.0\times 10^{-9}\ C} \) point charge.
▶️ Answer / Explanation
Use:
\( \mathrm{E = \dfrac{1}{4\pi\epsilon_0} \dfrac{Q}{r^2}} \)
\( \mathrm{E = 9\times10^{9} \cdot \dfrac{3.0\times10^{-9}}{(0.40)^2}} \)
\( \mathrm{E = 9\times10^{9} \cdot \dfrac{3.0\times10^{-9}}{0.16}} \)
\( \mathrm{E = 168.75\ N\,C^{-1}} \)
Electric field = \( \mathrm{1.7\times10^{2}\ N\,C^{-1}} \)
Example
A point charge produces an electric field of \( \mathrm{4500\ N\,C^{-1}} \) at a distance of 0.20 m. Calculate the value of the charge.
▶️ Answer / Explanation
Rearrange:
\( \mathrm{Q = E(4\pi\epsilon_0) r^2} \)
Or use \( \mathrm{E = \dfrac{1}{4\pi\epsilon_0} \dfrac{Q}{r^2}} \Rightarrow Q = \dfrac{Er^2}{9\times10^9} \)
\( \mathrm{Q = \dfrac{4500(0.20)^2}{9\times10^{9}}} \)
\( \mathrm{Q = \dfrac{4500 \cdot 0.04}{9\times10^{9}}} \)
\( \mathrm{Q = \dfrac{180}{9\times10^{9}} = 2.0\times10^{-8}\ C} \)
Charge = \( \mathrm{2.0\times10^{-8}\ C} \)
Example
A \( \mathrm{-5.0\times10^{-6}\ C} \) point charge creates an electric field. At what distance from the charge is the field strength equal to \( \mathrm{1.0\times10^{4}\ N\,C^{-1}} \)? State the direction of the field at that point.
▶️ Answer / Explanation
Use:
\( \mathrm{E = \dfrac{1}{4\pi\epsilon_0} \dfrac{Q}{r^2}} \)
Rearrange for distance:
\( \mathrm{r = \sqrt{\dfrac{1}{4\pi\epsilon_0}\dfrac{Q}{E}}} \)
Substitute values (use magnitude of Q):
\( \mathrm{r = \sqrt{\dfrac{9\times10^{9} \cdot 5.0\times10^{-6}}{1.0\times10^{4}}}} \)
\( \mathrm{r = \sqrt{\dfrac{45\times10^{3}}{10^{4}}}} \)
\( \mathrm{r = \sqrt{4.5}} = 2.12\ m \)
Direction:
- The charge is negative.
- Electric field points *toward* negative charges.
Distance = 2.12 m, field direction = toward the charge.