AP Physics 2- 10.3 Electric Fields- Study Notes- New Syllabus
AP Physics 2- 10.3 Electric Fields – Study Notes
AP Physics 2- 10.3 Electric Fields – Study Notes – per latest Syllabus.
Key Concepts:
- Electric Field due to Point Charges
- Electric Field Generated by Charged Conductors or Insulators
Electric Field due to Point Charges
The electric field (\( \vec{E} \)) at a point in space is the force experienced per unit positive charge placed at that point. It represents the region around a charge where it exerts an electric force.
\( \vec{E} = \dfrac{\vec{F}}{q_{test}} \)
For a point charge \( Q \):
\( E = \dfrac{1}{4\pi \varepsilon_0} \dfrac{|Q|}{r^2} \)
- Direction: Radially outward from \( Q \) if \( Q > 0 \), inward if \( Q < 0 \).
- Unit: Newton per Coulomb (N/C) or Volt per meter (V/m).
Net Electric Field (Superposition Principle):
If multiple charges are present, the total electric field is the vector sum of fields due to individual charges.
- Mathematically:
\( \vec{E}_{net} = \sum_i \vec{E}_i \)
- Direction must be carefully considered (vector nature).
Electric Field Line Diagrams:
- Field lines represent the direction of the electric field at different points.
- Properties of field lines:
- Originate from positive charges and terminate on negative charges.
- Never intersect each other.
- Density of lines ∝ field strength.
- Symmetry depends on charge configuration.
| Configuration | Electric Field Line Diagram |
|---|---|
| Single Point Charge: Radial lines (outward for +Q, inward for –Q). |  |
| Two Like Charges: Field lines repel each other (symmetrical pattern, lines bend away). |  |
| Two Unlike Charges: Field lines connect from + to – (dipole pattern). |  |
Example :
Find the net electric field at a point midway between two charges \( +2 \, \mu C \) and \( -2 \, \mu C \) placed 0.1 m apart.
▶️ Answer/Explanation
Step 1: Distance of midpoint from each charge = \( 0.05 \, m \).
Step 2: Electric field due to one charge: \( E = \dfrac{1}{4\pi \varepsilon_0} \dfrac{|Q|}{r^2} = (9 \times 10^9) \dfrac{2 \times 10^{-6}}{(0.05)^2} \)
= \( (9 \times 10^9)(2 \times 10^{-6})/2.5 \times 10^{-3} \)
= \( 7.2 \times 10^{6} \, N/C \)
Step 3: Directions → both fields point in the same direction (from + to – across the midpoint).
So, \( E_{net} = 2E = 1.44 \times 10^{7} \, N/C \).
Answer: Net electric field = \( 1.44 \times 10^{7} \, N/C \), directed from + to –.
Example :
Two equal positive charges \( +3 \, \mu C \) are placed 0.2 m apart. Find the net electric field at the midpoint between them.
▶️ Answer/Explanation
Step 1: Each charge produces an electric field of magnitude: \( E = (9 \times 10^9) \dfrac{3 \times 10^{-6}}{(0.1)^2} \)
= \( (9 \times 10^9)(3 \times 10^{-6})/0.01 \)
= \( 2.7 \times 10^{6} \, N/C \).
Step 2: At the midpoint, directions are opposite (both point away from each charge).
Thus, the fields cancel → \( E_{net} = 0 \).
Answer: Net electric field at midpoint = \( 0 \, N/C \).
Electric Field Generated by Charged Conductors or Insulators
Conductors:
In a conductor, free electrons move easily → charges redistribute themselves.
Key properties:
- Excess charge resides only on the surface of a conductor.
- Electric field inside a conductor in electrostatic equilibrium = zero.
- Outside the conductor, the electric field is perpendicular to the surface.
- For a spherical conductor (radius \(R\)):
\( E = \dfrac{1}{4\pi \varepsilon_0} \dfrac{Q}{r^2}, \, (r \geq R) \)
\( E = 0, \, (r < R) \)
- Thus, a charged spherical conductor produces the same field as if all charge were concentrated at its center.
Insulators (Dielectrics):
Charges are bound and cannot move freely.
- When charged, the charge distribution remains “frozen” at specific locations (surface or volume).
- Electric field depends on distribution:
- Uniformly charged line → produces a cylindrical field pattern.
- Uniformly charged sheet → produces a nearly constant electric field near the surface.
- Uniformly charged sphere → field inside increases linearly with radius, outside behaves like point charge.
- Inserting an insulator into an electric field → reduces the effective field because of polarization.
Field Line Behavior:
- For conductors: Field lines are normal (perpendicular) to surface, never tangential.
- For insulators: Field lines depend on fixed charge distribution and polarization effects.
Example :
A spherical conductor of radius 0.1 m carries a charge of \( +4 \, \mu C \). Find the electric field at (a) the surface, (b) 0.05 m inside, and (c) 0.2 m outside.
▶️ Answer/Explanation
(a) On surface: \( E = \dfrac{1}{4\pi \varepsilon_0} \dfrac{Q}{R^2} \)
= \( (9 \times 10^9) \dfrac{4 \times 10^{-6}}{(0.1)^2} \)
= \( 3.6 \times 10^6 \, N/C \).
(b) Inside (r < R): \( E = 0 \).
(c) Outside at 0.2 m: \( E = (9 \times 10^9) \dfrac{4 \times 10^{-6}}{(0.2)^2} \)
= \( 9.0 \times 10^5 \, N/C \).
Answer: (a) \( 3.6 \times 10^6 \, N/C \) (b) \( 0 \) (c) \( 9.0 \times 10^5 \, N/C \)