AP Physics C E&M- 8.3 Electric Fields- Study Notes- New Syllabus
AP Physics C E&M- 8.3 Electric Fields – Study Notes
AP Physics C E&M- 8.3 Electric Fields – Study Notes – per latest Syllabus.
Key Concepts:
- Electric Field Produced by a Charged Object or Point Charges
- Electric Field Generated by Charged Conductors and Insulators
Electric Field Produced by a Charged Object or Point Charges
The electric field is a vector field that describes the force experienced by a small positive test charge placed in the vicinity of a charged object. It represents how charges interact through space without direct contact.
Electric Field Due to a Point Charge:
The magnitude of the electric field produced by a single point charge is:
\( \mathrm{E = \dfrac{k_e |q|}{r^2}} \)
- \(\mathrm{E}\) = electric field strength (\(\mathrm{N/C}\) or \(\mathrm{V/m}\))
- \(\mathrm{q}\) = source charge
- \(\mathrm{r}\) = distance from the charge
- \(\mathrm{k_e = 8.99 \times 10^9 \, N \, m^2 / C^2}\)
The field direction is radially outward for positive charges and radially inward for negative charges.
Superposition Principle for Multiple Point Charges:
The net electric field at a point due to multiple charges is the vector sum of the fields due to each individual charge:
\( \mathrm{\vec{E}_{net} = \sum_i \vec{E}_i = \sum_i \dfrac{k_e q_i}{r_i^2} \hat{r}_i} \)
- \(\mathrm{q_i}\) = i-th charge
- \(\mathrm{r_i}\) = distance from i-th charge to the point of interest
- \(\mathrm{\hat{r}_i}\) = unit vector pointing from the charge to the point
Field Lines:
- Point outward from positive charges and inward toward negative charges.
- The density of field lines represents the field strength.
- Field lines never cross.
Units: Electric field is measured in newtons per coulomb (\(\mathrm{N/C}\)) or equivalently in volts per meter (\(\mathrm{V/m}\)).
Example :
Find the electric field magnitude and direction at a point \(0.2 \, m\) away from a point charge of \(+3.0 \, \mu C\).
▶️ Answer/Explanation
Step 1: Use the formula: \( \mathrm{E = \dfrac{k_e |q|}{r^2}} \)
Step 2: Substitute: \( \mathrm{E = \dfrac{(8.99 \times 10^9)(3.0 \times 10^{-6})}{(0.2)^2}} \)
Step 3: \( \mathrm{E \approx 6.74 \times 10^5 \, N/C} \)
Step 4: Since the charge is positive, the electric field points radially outward.
Final Answer: \( \mathrm{E = 6.7 \times 10^5 \, N/C} \), outward from the charge.
Example :
Two charges, \(+2.0 \, \mu C\) and \(-2.0 \, \mu C\), are separated by \(0.4 \, m\). Find the net electric field at the midpoint between them.
▶️ Answer/Explanation
Step 1: Distance from midpoint to each charge = \(0.2 \, m\).
Step 2: Field due to each charge: \( \mathrm{E = \dfrac{k_e |q|}{r^2} = \dfrac{(8.99 \times 10^9)(2.0 \times 10^{-6})}{(0.2)^2}} \)
\( \mathrm{E \approx 4.5 \times 10^5 \, N/C} \)
Step 3: At the midpoint, the fields from each charge point in the same direction (toward the negative charge).
Step 4: Net field: \( \mathrm{E_{net} = 2E = 9.0 \times 10^5 \, N/C} \), directed toward the negative charge.
Final Answer: \( \mathrm{E_{net} = 9.0 \times 10^5 \, N/C} \), directed from positive toward negative.
| 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). |  |
Electric Field Generated by Charged Conductors and Insulators
The behavior of electric fields depends strongly on whether the charge is placed on a conductor (where charges move freely) or an insulator (where charges are fixed in place).
1. Conductors:
In conductors, free electrons move until the system reaches electrostatic equilibrium.
- In equilibrium:
- The electric field inside a conductor is zero.
- Excess charge resides entirely on the outer surface of the conductor.
- The electric field at the surface is perpendicular to the surface.
- The field just outside the surface is related to the surface charge density \(\mathrm{\sigma}\): \( \mathrm{E = \dfrac{\sigma}{\varepsilon_0}} \)
2. Insulators:
In insulators (dielectrics), charges are fixed in place and cannot move freely.
- The electric field is produced by the actual distribution of charges within the material.
- Unlike conductors, excess charge can remain in the interior of the material.
- The field inside depends on how the charges are distributed (uniform, point, or polarized).
Comparison of Fields:
- Conductors: Field exists only outside; inside field is zero at equilibrium.
- Insulators: Field can exist both inside and outside, determined by fixed charge locations.
Units: The electric field is measured in newtons per coulomb (\(\mathrm{N/C}\)) or volts per meter (\(\mathrm{V/m}\)).
Example :
A spherical conductor of radius \(0.1 \, m\) carries a total charge of \(+4.0 \, \mu C\). Find the electric field just outside its surface.
▶️ Answer/Explanation
Step 1: Surface charge density: \( \mathrm{\sigma = \dfrac{Q}{4 \pi R^2}} \)
\( \mathrm{\sigma = \dfrac{4.0 \times 10^{-6}}{4 \pi (0.1)^2} \approx 3.18 \times 10^{-5} \, C/m^2} \)
Step 2: Electric field just outside: \( \mathrm{E = \dfrac{\sigma}{\varepsilon_0} = \dfrac{3.18 \times 10^{-5}}{8.85 \times 10^{-12}} \approx 3.6 \times 10^6 \, N/C} \)
Final Answer: \( \mathrm{E \approx 3.6 \times 10^6 \, N/C} \), directed radially outward.
Example:
A non-conducting sphere of radius \(0.1 \, m\) has a uniform charge distribution with total charge \(+4.0 \, \mu C\). Find the electric field at a point \(0.05 \, m\) from its center (inside the sphere).
▶️ Answer/Explanation
Step 1: For a uniformly charged insulating sphere, field inside: \( \mathrm{E = \dfrac{1}{4 \pi \varepsilon_0} \dfrac{Q r}{R^3}} \)
Step 2: Substitute: \( \mathrm{E = (9 \times 10^9) \dfrac{(4.0 \times 10^{-6})(0.05)}{(0.1)^3}} \)
\( \mathrm{E \approx 1.8 \times 10^6 \, N/C} \)
Final Answer: \( \mathrm{E = 1.8 \times 10^6 \, N/C} \), directed radially outward.