CIE AS/A Level Physics 18.3 Electric force between point charges Study Notes- 2025-2027 Syllabus
CIE AS/A Level Physics 18.3 Electric force between point charges Study Notes – New Syllabus
CIE AS/A Level Physics 18.3 Electric force between point charges 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:
- understand that, for a point outside a spherical conductor, the charge on the sphere may be considered to be a point charge at its centre
- recall and use Coulomb’s law F = Q1Q2 / (4πε0 r 2) for the force between two point charges in free space
Electric Field Outside a Charged Spherical Conductor
When a conductor is shaped as a sphere and carries charge, all the charge distributes itself uniformly over the surface. For any point located outside the sphere, the sphere behaves electrically as if:
All of its charge were concentrated at a single point at its centre.
This means the electric field and force outside a charged spherical conductor are identical to those produced by a point charge located at the sphere’s centre with the same total charge.
Key Consequences:
- Coulomb’s law may be applied by treating the sphere as a point charge.
- The electric field outside the sphere is radial (pointing away if positively charged, toward if negatively charged).
- Field strength outside the sphere is given by:
\( \mathrm{E = \dfrac{1}{4\pi\epsilon_0} \dfrac{Q}{r^2}} \)
- Where \( \mathrm{r} \) is the distance from the centre of the sphere.
- This is identical to the expression for a point charge.
Why this works:
Due to symmetry, the contributions from each surface element combine to produce an electric field identical to a point charge at the centre for all points external to the sphere.
Example
You stand 2 m away from a metal sphere carrying a charge of \( \mathrm{+5\,\mu C} \). Explain how to treat the sphere when calculating the electric field at your position.
▶️ Answer / Explanation
Since you are outside the sphere, its charge can be considered to be concentrated at its centre. So the sphere acts like a point charge of \( \mathrm{+5\,\mu C} \) located at its centre. You can apply the point-charge formula for the electric field.
Example
A spherical conductor of radius 0.30 m carries a charge of \( \mathrm{2.0\times 10^{-6}\ C} \). Calculate the electric field strength at a point 0.80 m from its centre.
▶️ Answer / Explanation
Since 0.80 m > 0.30 m, the point is outside the sphere → treat it as a point charge.
\( \mathrm{E = \dfrac{1}{4\pi\epsilon_0} \dfrac{Q}{r^2}} \)
\( \mathrm{E = 9\times 10^{9} \cdot \dfrac{2.0\times 10^{-6}}{(0.80)^2}} \)
\( \mathrm{E = 9\times 10^{9} \cdot \dfrac{2.0\times 10^{-6}}{0.64}} \)
\( \mathrm{E = 2.81\times 10^{4}\ N\,C^{-1}} \)
Electric field ≈ \( \mathrm{2.8\times 10^{4}\ N\,C^{-1}} \)
Example
A spherical conductor of radius 0.10 m carries a charge of \( \mathrm{-6.0\times 10^{-6}\ C} \). A small \( \mathrm{+3.0\times 10^{-9}\ C} \) test charge is placed at a distance of 0.25 m from the sphere’s centre. Calculate the force on the test charge.
▶️ Answer / Explanation
Since 0.25 m > 0.10 m, the test charge is outside → treat sphere as a point charge at its centre.
Step 1: Electric field at test charge position
\( \mathrm{E = \dfrac{1}{4\pi\epsilon_0} \dfrac{Q}{r^2}} \)
\( \mathrm{E = 9\times 10^{9} \cdot \dfrac{-6.0\times 10^{-6}}{(0.25)^2}} \)
\( \mathrm{E = 9\times 10^{9} \cdot \dfrac{-6.0\times 10^{-6}}{0.0625}} \)
\( \mathrm{E = -8.64\times 10^{5}\ N\,C^{-1}} \)
(Negative → direction is toward sphere because sphere is negatively charged.)
Step 2: Force on test charge
\( \mathrm{F = qE = (3.0\times 10^{-9})(-8.64\times 10^{5})} \)
\( \mathrm{F = -2.59\times 10^{-3}\ N} \)
Force magnitude = \( \mathrm{2.6\times 10^{-3}\ N} \)
Direction: toward sphere (negative sphere attracts positive test charge).
Coulomb’s Law for Force Between Two Point Charges
Coulomb’s law gives the electric force between two point charges in free space (vacuum). It states that the force is:
- Directly proportional to the product of the charges
- Inversely proportional to the square of the distance between them
- Along the line joining the two charges
Coulomb’s Law Formula:
\( \mathrm{F = \dfrac{Q_1 Q_2}{4\pi \epsilon_0 r^2}} \)
- \( \mathrm{F} \) = electric force (N)
- \( \mathrm{Q_1, Q_2} \) = charges (C)
- \( \mathrm{r} \) = separation between charges (m)
- \( \mathrm{\epsilon_0 = 8.85\times 10^{-12}\ F\,m^{-1}} \)
- \( \mathrm{\dfrac{1}{4\pi\epsilon_0} = 9.0\times 10^9\ N\,m^2\,C^{-2}} \)
Nature of Force:
- Like charges repel.
- Unlike charges attract.
- The force acts along the line joining the charges.
Example
Two charges, \( \mathrm{Q_1 = +2.0\times10^{-6}\ C} \) and \( \mathrm{Q_2 = +3.0\times10^{-6}\ C} \), are separated by 0.50 m. Calculate the force between them.
▶️ Answer / Explanation
Use:
\( \mathrm{F = \dfrac{1}{4\pi\epsilon_0} \dfrac{Q_1Q_2}{r^2}} \)
Substitute:
\( \mathrm{F = 9\times10^{9} \cdot \dfrac{(2.0\times10^{-6})(3.0\times10^{-6})}{(0.50)^2}} \)
\( \mathrm{F = 9\times10^{9} \cdot \dfrac{6.0\times10^{-12}}{0.25}} \)
\( \mathrm{F = 2.16\times10^{-1}\ N} \)
Force = 0.216 N (repulsive)
Example
Two charges, \( \mathrm{+5.0\times10^{-6}\ C} \) and \( \mathrm{-4.0\times10^{-6}\ C} \), are placed 0.20 m apart. Find the magnitude and nature of the force between them.
▶️ Answer / Explanation
Magnitude:
\( \mathrm{F = 9\times10^{9} \cdot \dfrac{(5.0\times10^{-6})(4.0\times10^{-6})}{(0.20)^2}} \)
\( \mathrm{F = 9\times10^{9} \cdot \dfrac{20\times10^{-12}}{0.04}} \)
\( \mathrm{F = 4.5\ N} \)
Nature of force: charges are opposite → attractive.
Example
Three charges lie on a straight line. A charge \( \mathrm{Q_1 = +6.0\times10^{-6}\ C} \) is at x = 0. A charge \( \mathrm{Q_2 = -3.0\times10^{-6}\ C} \) is at x = 0.30 m. A charge \( \mathrm{Q_3 = +2.0\times10^{-6}\ C} \) is at x = 0.80 m.
Calculate the net force on \( \mathrm{Q_2} \).
▶️ Answer / Explanation
Force on \( Q_2 \) due to \( Q_1 \):
Distance = 0.30 m
\( \mathrm{F_{21} = 9\times10^{9} \cdot \dfrac{(6.0\times10^{-6})(3.0\times10^{-6})}{(0.30)^2}} \)
\( \mathrm{F_{21} = 9\times10^{9} \cdot \dfrac{18\times10^{-12}}{0.09}} \)
\( \mathrm{F_{21} = 1.8\ N} \)
Since \( Q_1 \) is positive and \( Q_2 \) is negative → force is attractive → to the left.
Force on \( Q_2 \) due to \( Q_3 \):
Distance = 0.50 m
\( \mathrm{F_{23} = 9\times10^{9} \cdot \dfrac{(3.0\times10^{-6})(2.0\times10^{-6})}{(0.50)^2}} \)
\( \mathrm{F_{23} = 9\times10^{9} \cdot \dfrac{6\times10^{-12}}{0.25}} \)
\( \mathrm{F_{23} = 0.216\ N} \)
Since \( Q_3 \) is positive and \( Q_2 \) is negative → attractive → to the right.
Net force on \( Q_2 \):
Left force: 1.8 N Right force: 0.216 N
Net = \( \mathrm{1.8 – 0.216 = 1.584\ N} \) to the left
Net force = \( \mathrm{1.58\ N} \) to the left.