AP Physics 2- 10.5 Electric Potential- Study Notes- New Syllabus
AP Physics 2- 10.5 Electric Potential – Study Notes
AP Physics 2- 10.5 Electric Potential – Study Notes – per latest Syllabus.
Key Concepts:
- Electric Potential(V)
- Electric Potential Difference (ΔV)
- Relationship Between Electric Potential (V) and Electric Field (E)
- Equipotential Lines
Electric Potential (V)
Electric potential describes the electric potential energy per unit charge at a point in space. It is a scalar quantity, measured in volts (V), where \( 1 \, \text{V} = 1 \, \text{J/C} \).
Key Equation (Point Charge):
\( \text{V} = \dfrac{k \text{Q}}{r} \)
- \( \text{Q} \): Charge producing the potential
- \( r \): Distance from the charge to the point
- \( k \): Coulomb constant \( = 9.0 \times 10^9 \, \text{N·m²/C²} \)
Superposition Principle:
The total electric potential due to multiple point charges is the algebraic sum of the potentials due to each individual charge:
\( \text{V}_{\text{total}} = \sum_i \dfrac{k \text{Q}_i}{r_i} \)
Since electric potential is a scalar, we add algebraically rather than vectorially.
Example :
Two point charges, \( \text{Q}_1 = 2 \, \mu\text{C} \) and \( \text{Q}_2 = -3 \, \mu\text{C} \), are located 0.5 m and 0.3 m from a point P. Find the electric potential at P.
▶️ Answer/Explanation
\( \text{V}_{\text{total}} = k \dfrac{\text{Q}_1}{r_1} + k \dfrac{\text{Q}_2}{r_2} \)
= \( (9.0 \times 10^9) \dfrac{2 \times 10^{-6}}{0.5} + (9.0 \times 10^9) \dfrac{-3 \times 10^{-6}}{0.3} \)
= \( 3.6 \times 10^4 – 9.0 \times 10^4 \)
= \( -5.4 \times 10^4 \, \text{V} \)
Answer: Electric potential at P ≈ \(-5.4 \times 10^4 \, \text{V}\)
Electric Potential Difference (ΔV)
The electric potential difference between two points is the change in electric potential energy per unit charge when a test charge is moved between the two points.
- It is a scalar quantity, measured in Volts (V).
Key Equation:
\( \Delta V = \dfrac{\Delta U_E}{q} \)
- \( \Delta V \): Potential difference (V)
- \( \Delta U_E \): Change in electric potential energy (J)
- \( q \): Test charge (C)
Additional Notes:
- Electric potential difference may also result from chemical processes that cause positive and negative charges to separate (e.g., in a battery).
- When conductors are in electrical contact, electrons redistribute until the surfaces of the conductors are at the same potential.
Example :
A test charge of \( q = 2 \, C \) experiences a change in potential energy of \( \Delta U_E = 12 \, J \) when moved between two points. Find the potential difference.
▶️ Answer/Explanation
Formula: \( \Delta V = \dfrac{\Delta U_E}{q} \)
= \( \dfrac{12}{2} \)
= \( 6 \, V \)
Answer: The potential difference between the two points is \( 6 \, V \).
Relationship Between Electric Potential (V) and Electric Field (E)
The electric field is related to the rate of change of electric potential with distance. Electric field points in the direction of maximum decrease of potential.
- This means: charges naturally move from regions of high potential to low potential.
Key Equations:
For a uniform electric field:
\( E = – \dfrac{\Delta V}{d} \)
where \( d \) is the displacement in the direction of the field.
In general (1D case):
\( E_x = – \dfrac{dV}{dx} \)
Vector form (3D):
\( \vec{E} = – \nabla V \)
Properties:
- Electric field lines are always directed perpendicular to equipotential surfaces.
- A larger potential difference over a shorter distance → stronger electric field.
- No work is done when moving a charge along an equipotential surface (ΔV = 0, hence \(E \cdot d = 0\)).
Example :
Two parallel plates are separated by 0.02 m and connected to a 200 V battery. Find the electric field between the plates.
▶️ Answer/Explanation
Formula: \( E = – \dfrac{\Delta V}{d} \)
= \( – \dfrac{200}{0.02} \)
= \( -1.0 \times 10^{4} \, N/C \)
Answer: Electric field = \( 1.0 \times 10^{4} \, N/C \), directed from the positive plate to the negative plate.
Equipotential Lines
Equipotential lines (or surfaces in 3D) are lines connecting points that have the same electric potential. No work is required to move a charge along an equipotential line because ΔV = 0 along the path.
Key Properties:
- Electric field lines are always perpendicular to equipotential lines.
- Equipotential lines never intersect.
- The spacing of equipotential lines indicates the strength of the electric field:
- Closer lines → stronger field.
- Farther apart → weaker field.
- For a point charge:
- Equipotential surfaces are concentric spheres (or circles in 2D) around the charge.
- For a uniform electric field:
- Equipotential surfaces are parallel planes perpendicular to the field lines.
- For multiple charges: Equipotential lines are distorted according to superposition of potentials.
Example :
Draw equipotential lines for a single positive point charge and describe the electric field direction.
▶️ Answer/Explanation
Equipotential lines are concentric circles centered at the point charge.
The electric field is radial, pointing outward from the positive charge, and perpendicular to all equipotential lines.
Example :
Two parallel plates connected to a battery have a uniform electric field between them. Draw the equipotential lines.
▶️ Answer/Explanation
Equipotential lines are equally spaced parallel lines (planes in 3D) perpendicular to the field lines.
Work done moving a charge along any line = 0.