CIE AS/A Level Physics 18.5 Electric potential Study Notes- 2025-2027 Syllabus
CIE AS/A Level Physics 18.5 Electric potential Study Notes – New Syllabus
CIE AS/A Level Physics 18.5 Electric potential 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:
- define electric potential at a point as the work done per unit positive charge in bringing a small test charge from infinity to the point
- recall and use the fact that the electric field at a point is equal to the negative of potential gradient at that point
- use V = Q / (4πε0r) for the electric potential in the field due to a point charge
- understand how the concept of electric potential leads to the electric potential energy of two point charges and use EP = Qq / (4πε0 r)
Electric Potential at a Point
Electric potential describes how much electric potential energy a positive charge would have at a point in an electric field.
Definition:
Electric potential at a point is the work done per unit positive charge in bringing a small test charge from infinity to that point.
This means:
- Infinity is the reference point where electric potential is taken as zero.
- A positive test charge is used by convention.
- The work is done against or by the electric field depending on the charge creating the field.
- Electric potential is a scalar quantity.
Formula:
\( \mathrm{V = \dfrac{W}{q}} \)
- \( \mathrm{V} \) = electric potential (V or J C⁻¹)
- \( \mathrm{W} \) = work done in bringing the charge from infinity (J)
- \( \mathrm{q} \) = test charge (C)
Key Points:
- A positive source charge creates a region of positive potential.
- A negative source charge creates a negative potential.
- Closer to the charge → larger magnitude of potential.
Example
It takes \( \mathrm{0.12\ J} \) of work to bring a \( \mathrm{4.0\times10^{-6}\ C} \) test charge from infinity to a point. Calculate the electric potential at that point.
▶️ Answer / Explanation
Use:
\( \mathrm{V = \dfrac{W}{q}} \)
\( \mathrm{V = \dfrac{0.12}{4.0\times10^{-6}} = 3.0\times10^{4}\ V} \)
Electric potential = \( \mathrm{3.0\times10^{4}\ V} \)
Example
If the electric potential at a point is \( \mathrm{5000\ V} \), how much work is required to bring a \( \mathrm{2.0\times10^{-6}\ C} \) charge from infinity to that point?
▶️ Answer / Explanation
Use:
\( \mathrm{W = qV} \)
\( \mathrm{W = (2.0\times10^{-6})(5000)} \)
\( \mathrm{W = 0.01\ J} \)
Work required = 0.01 J
Example
Bringing a \( \mathrm{+1.0\times10^{-9}\ C} \) test charge from infinity to point A requires \( \mathrm{-2.5\times10^{-3}\ J} \) of work (the field does the work). Determine the electric potential at point A and explain the sign.
▶️ Answer / Explanation
Use:
\( \mathrm{V = \dfrac{W}{q}} \)
\( \mathrm{V = \dfrac{-2.5\times10^{-3}}{1.0\times10^{-9}}} \)
\( \mathrm{V = -2.5\times10^{6}\ V} \)
Electric potential = \( \mathrm{-2.5\times10^{6}\ V} \)
Meaning:
The negative sign shows that the point is in the field of a negative source charge— work is done by the field when a positive test charge moves closer.
Electric Field and Potential Gradient
The electric field at a point in space is related to how rapidly the electric potential changes with distance. This relationship is fundamental in electrostatics.
Key Fact:
\( \mathrm{E = -\dfrac{dV}{dr}} \)
- \( \mathrm{E} \) = electric field strength (V m⁻¹ or N C⁻¹)
- \( \mathrm{\dfrac{dV}{dr}} \) = potential gradient (rate of change of potential with distance)
- Negative sign indicates field points in direction of decreasing potential
Meaning:
The electric field at a point equals the rate at which electric potential decreases with distance from that point.
Physical Interpretation:
- Steep decrease in potential → strong electric field.
- Slow or flat potential → weak electric field.
- Electric field points from high potential to low potential.
Special Case — Uniform Electric Field:
If the potential changes linearly between parallel plates, \( \mathrm{E = -\dfrac{\Delta V}{\Delta d}} \).
Example
The electric potential decreases uniformly from 200 V to 100 V over a distance of 0.50 m. Find the electric field strength.
▶️ Answer / Explanation
Potential gradient:
\( \mathrm{\dfrac{\Delta V}{\Delta d} = \dfrac{100 – 200}{0.50} = -200\ V\,m^{-1}} \)
Electric field:
\( \mathrm{E = -\dfrac{dV}{dr} = -(-200) = 200\ V\,m^{-1}} \)
Electric field = \( \mathrm{200\ V\,m^{-1}} \)
Example
The graph of electric potential vs distance has a slope of \( \mathrm{-5000\ V\,m^{-1}} \) at a point. Determine the electric field there.
▶️ Answer / Explanation
The slope of the graph is the potential gradient:
\( \mathrm{\dfrac{dV}{dr} = -5000\ V\,m^{-1}} \)
Electric field:
\( \mathrm{E = -\dfrac{dV}{dr} = -(-5000) = 5000\ V\,m^{-1}} \)
Electric field = \( \mathrm{5.0\times10^{3}\ V\,m^{-1}} \)
Example
The electric potential outside a point charge varies according to \( \mathrm{V = \dfrac{1}{4\pi\epsilon_0}\dfrac{Q}{r}} \). Derive an expression for the electric field and calculate its value at \( \mathrm{r = 0.30\ m} \) for \( \mathrm{Q = +8.0\times10^{-6}\ C} \).
▶️ Answer / Explanation
Step 1: Differentiate potential
\( \mathrm{V = \dfrac{kQ}{r}} \Rightarrow \dfrac{dV}{dr} = -\dfrac{kQ}{r^2}} \)
Step 2: Electric field
\( \mathrm{E = -\dfrac{dV}{dr} = \dfrac{kQ}{r^2}} \)
(This matches the standard point-charge field.)
Step 3: Substitute values
\( \mathrm{E = \dfrac{9\times10^{9} \cdot 8.0\times10^{-6}}{(0.30)^2}} \)
\( \mathrm{E = \dfrac{72\times10^{3}}{0.09}} \)
\( \mathrm{E = 8.0\times10^{5}\ V\,m^{-1}} \)
Electric field = \( \mathrm{8.0\times10^{5}\ V\,m^{-1}} \)
Electric Potential Due to a Point Charge
The electric potential at a distance \( \mathrm{r} \) from a point charge in free space is given by:
\( \mathrm{V = \dfrac{Q}{4\pi\epsilon_0 r}} \)
- \( \mathrm{V} \) = electric potential (V or J C⁻¹)
- \( \mathrm{Q} \) = point charge (C)
- \( \mathrm{r} \) = distance from the charge (m)
- \( \mathrm{\epsilon_0 = 8.85\times10^{-12}\ F\,m^{-1}} \)
Key Features:
- Electric potential decreases as distance increases → inverse relationship.
- Positive charge produces positive potential; negative charge produces negative potential.
- Potential is a scalar quantity.
- Reference point: \( \mathrm{V = 0} \) at infinity.
Physical Meaning:
The potential at distance \( \mathrm{r} \) is the work done per unit positive charge in bringing the charge from infinity to that point.
Example
Find the electric potential 0.50 m away from a point charge \( \mathrm{Q = +4.0\times10^{-6}\ C} \).
▶️ Answer / Explanation
Use:
\( \mathrm{V = \dfrac{1}{4\pi\epsilon_0}\dfrac{Q}{r}} \)
\( \mathrm{V = 9\times10^{9}\cdot \dfrac{4.0\times10^{-6}}{0.50}} \)
\( \mathrm{V = 9\times10^{9}\cdot 8.0\times10^{-6}} \)
\( \mathrm{V = 7.2\times10^{4}\ V} \)
Electric potential = \( \mathrm{7.2\times10^{4}\ V} \)
Example
The electric potential at a point 0.30 m from a point charge is \( \mathrm{1.5\times10^{4}\ V} \). Find the charge.
▶️ Answer / Explanation
Rearrange:
\( \mathrm{Q = \dfrac{Vr}{(1/(4\pi\epsilon_0))}} = \dfrac{Vr}{9\times10^9} \)
Substitute:
\( \mathrm{Q = \dfrac{(1.5\times10^{4})(0.30)}{9\times10^{9}}} \)
\( \mathrm{Q = \dfrac{4500}{9\times10^{9}} = 5.0\times10^{-7}\ C} \)
Charge = \( \mathrm{5.0\times10^{-7}\ C} \)
Example
A point charge produces a potential of \( \mathrm{-3.0\times10^{3}\ V} \) at a distance of 0.20 m. Determine the charge and state whether it is positive or negative.
▶️ Answer / Explanation
Use:
\( \mathrm{Q = Vr(4\pi\epsilon_0)} \)
Or using k = \( \mathrm{9\times10^9} \):
\( \mathrm{Q = \dfrac{Vr}{k}} \)
Substitute:
\( \mathrm{Q = \dfrac{(-3.0\times10^{3})(0.20)}{9\times10^{9}}} \)
\( \mathrm{Q = \dfrac{-600}{9\times10^{9}} = -6.7\times10^{-8}\ C} \)
The charge is negative because the potential is negative.
Charge = \( \mathrm{-6.7\times10^{-8}\ C} \)
Electric Potential Energy of Two Point Charges
Electric potential energy is the energy stored in a system of charges due to their electric interaction. It arises directly from the concept of electric potential.
Link Between Potential and Potential Energy:
If a charge \( \mathrm{q} \) is placed at a point where the electric potential is \( \mathrm{V} \), its electric potential energy is \( \mathrm{E_P = qV} \).
For the electric potential created by another point charge \( \mathrm{Q} \):
\( \mathrm{V = \dfrac{Q}{4\pi\epsilon_0 r}} \)
Substituting this into \( \mathrm{E_P = qV} \):
\( \mathrm{E_P = \dfrac{Qq}{4\pi\epsilon_0 r}} \)
- \( \mathrm{E_P} \) = electric potential energy (J)
- \( \mathrm{Q} \), \( \mathrm{q} \) = interacting charges (C)
- \( \mathrm{r} \) = separation (m)
- \( \epsilon_0 \) = permittivity of free space
Sign of Electric Potential Energy:
- If both charges have the same sign → \( \mathrm{E_P > 0} \) (repulsive interaction).
- If charges have opposite signs → \( \mathrm{E_P < 0} \) (attractive interaction).
- The greater the separation, the closer \( \mathrm{E_P} \) is to zero.
Physical Meaning:
Electric potential energy is the work required to bring two charges from infinite separation to a separation of \( \mathrm{r} \).
Example
Calculate the electric potential energy between \( \mathrm{Q = +2.0\times10^{-6}\ C} \) and \( \mathrm{q = +3.0\times10^{-6}\ C} \) separated by 0.50 m.
▶️ Answer / Explanation
Use:
\( \mathrm{E_P = \dfrac{Qq}{4\pi\epsilon_0 r}} \)
\( \mathrm{E_P = 9\times10^{9} \cdot \dfrac{(2\times10^{-6})(3\times10^{-6})}{0.50}} \)
\( \mathrm{E_P = 9\times10^{9} \cdot 12\times10^{-12}} \)
\( \mathrm{E_P = 0.108\ J} \)
The potential energy = \( \mathrm{0.108\ J} \) (positive, repulsive).
Example
Two charges, \( \mathrm{Q = +8.0\times10^{-9}\ C} \) and \( \mathrm{q = -4.0\times10^{-9}\ C} \), are 0.25 m apart. Find the electric potential energy of the system and interpret the sign.
▶️ Answer / Explanation
Use:
\( \mathrm{E_P = 9\times10^{9} \cdot \dfrac{(8.0\times10^{-9})(4.0\times10^{-9})}{0.25}} \)
\( \mathrm{E_P = 9\times10^{9} \cdot \dfrac{32\times10^{-18}}{0.25}} \)
\( \mathrm{E_P = 9\times10^{9} \cdot 128\times10^{-18}} \)
\( \mathrm{E_P = -1.15\times10^{-6}\ J} \)
Negative sign → attractive interaction.
The two charges lower their energy by being close.
Example
An electron (charge \( \mathrm{-1.6\times10^{-19}\ C} \)) is placed \( \mathrm{3.0\times10^{-10}\ m} \) from a proton (charge \( \mathrm{+1.6\times10^{-19}\ C} \)). Calculate the electric potential energy of the electron–proton system.
▶️ Answer / Explanation
Use:
\( \mathrm{E_P = \dfrac{1}{4\pi\epsilon_0} \dfrac{Qq}{r}} \)
\( \mathrm{E_P = 9\times10^{9} \cdot \dfrac{(1.6\times10^{-19})(1.6\times10^{-19})}{3.0\times10^{-10}}} \)
\( \mathrm{E_P = 9\times10^{9} \cdot \dfrac{2.56\times10^{-38}}{3.0\times10^{-10}}} \)
\( \mathrm{E_P = 9\times10^{9} \cdot 8.53\times10^{-29}} \)
\( \mathrm{E_P = 7.68\times10^{-19}\ J} \)
Because charges are opposite, potential energy is negative:
\( \mathrm{E_P = -7.68\times10^{-19}\ J} \)
Electric potential energy = \( \mathrm{-7.7\times10^{-19}\ J} \)