AP Physics C E&M- 8.4 Electric Fields Charge Distributions- Study Notes- New Syllabus
AP Physics C E&M- 8.4 Electric Fields Charge Distributions – Study Notes
AP Physics C E&M- 8.4 Electric Fields Charge Distributions – Study Notes – per latest Syllabus.
Key Concepts:
- Electric Field Resulting from a Given Charge Distribution
Electric Field Resulting from a Given Charge Distribution
The electric field at a point in space due to a charge distribution is the vector sum of contributions from all individual charges that make up the distribution. Depending on whether the charge is localized or spread out, the distribution can be discrete (point charges) or continuous (line, surface, or volume charges).
Superposition Principle (Discrete Charges):
For a system of discrete point charges:
\( \mathrm{\vec{E}_{net} = \sum_i \vec{E}_i = \sum_i \dfrac{k_e q_i}{r_i^2} \hat{r}_i} \)
Continuous Charge Distributions (Integral Form):
- When charges are continuously distributed, the electric field is found by integration:
1. Line Charge:
For linear charge density \(\mathrm{\lambda = \dfrac{dq}{dl}}\):
\( \mathrm{\vec{E} = \dfrac{1}{4 \pi \varepsilon_0} \int \dfrac{\lambda \, dl}{r^2} \, \hat{r}} \)
2. Surface Charge:
For surface charge density \(\mathrm{\sigma = \dfrac{dq}{dA}}\):
\( \mathrm{\vec{E} = \dfrac{1}{4 \pi \varepsilon_0} \int \dfrac{\sigma \, dA}{r^2} \, \hat{r}} \)
3. Volume Charge:
For volume charge density \(\mathrm{\rho = \dfrac{dq}{dV}}\):
\( \mathrm{\vec{E} = \dfrac{1}{4 \pi \varepsilon_0} \int \dfrac{\rho \, dV}{r^2} \, \hat{r}} \)
Symmetry and Gauss’s Law(Topic – 8.6):
For highly symmetric charge distributions (spherical, cylindrical, planar), Gauss’s law provides a more direct method:
\( \mathrm{\oint \vec{E} \cdot d\vec{A} = \dfrac{Q_{enclosed}}{\varepsilon_0}} \)
Key Idea: For complex or asymmetric charge distributions, integration is required. For symmetric distributions, Gauss’s law greatly simplifies the calculation.
Example:
A thin rod of length \(L = 2.0 \, m\) carries a uniform charge of \(+6.0 \, \mu C\). Find the linear charge density \(\lambda\).
▶️ Answer/Explanation
Step 1: Use definition: \( \mathrm{\lambda = \dfrac{Q}{L}} \)
Step 2: Substitute values: \( \mathrm{\lambda = \dfrac{6.0 \times 10^{-6}}{2.0} = 3.0 \times 10^{-6} \, C/m} \)
Final Answer: \(\mathrm{\lambda = 3.0 \times 10^{-6} \, C/m}\).
Example:
A large, flat insulating sheet has a uniform surface charge density \(\mathrm{\sigma = 5.0 \times 10^{-6} \, C/m^2}\). Find the electric field just outside the surface.
▶️ Answer/Explanation
Step 1: For an infinite sheet of charge, the field is given by: \( \mathrm{E = \dfrac{\sigma}{2 \varepsilon_0}} \)
Step 2: Substitute: \( \mathrm{E = \dfrac{5.0 \times 10^{-6}}{2 \times 8.85 \times 10^{-12}}} \)
Step 3: \( \mathrm{E \approx 2.82 \times 10^5 \, N/C} \), directed perpendicular to the sheet.
Final Answer: \(\mathrm{E = 2.8 \times 10^5 \, N/C}\).