AP Physics C E&M- 9.1 Electric Potential Energy- Study Notes- New Syllabus
AP Physics C E&M- 9.1 Electric Potential Energy – Study Notes
AP Physics C E&M- 9.1 Electric Potential Energy – Study Notes – per latest Syllabus.
Key Concepts:
- Electric Potential Energy of a System
Electric Potential Energy of a System
Electric potential energy is the energy stored in a system of charged objects due to their relative positions. It represents the work required to assemble the system by bringing charges from infinity to their positions in the presence of electric forces
For Two Point Charges:
The potential energy of a pair of charges is:
\( \mathrm{U = \dfrac{1}{4 \pi \varepsilon_0} \dfrac{q_1 q_2}{r}} \)
- \(\mathrm{q_1, q_2}\) = charges
- \(\mathrm{r}\) = separation between charges
- \(\mathrm{\varepsilon_0 = 8.85 \times 10^{-12} \, C^2 / (N \, m^2)}\)
Sign Convention:
- If \(\mathrm{q_1 q_2 > 0}\) (like charges): \(U > 0\), repulsive interaction.
- If \(\mathrm{q_1 q_2 < 0}\) (unlike charges): \(U < 0\), attractive interaction.
System of Multiple Charges:
The total potential energy is the sum over all unique pairs of charges:
\( \mathrm{U_{total} = \dfrac{1}{4 \pi \varepsilon_0} \sum_{i<j} \dfrac{q_i q_j}{r_{ij}}} \)
Relation to Work:
- The work done by an external agent in assembling the system from infinity equals the potential energy stored in the system.
- If a charge is moved in the electric field, the change in potential energy is equal to the negative of the work done by the field.
Key Features:
- Potential energy depends only on relative positions of charges, not on their path of assembly.
- For continuous charge distributions, integration is required.
Example :
Two point charges, \(q_1 = +2.0 \, \mu C\) and \(q_2 = -3.0 \, \mu C\), are separated by \(0.5 \, m\). Find the potential energy of the system.
▶️ Answer/Explanation
Step 1: Use formula: \( \mathrm{U = \dfrac{1}{4 \pi \varepsilon_0} \dfrac{q_1 q_2}{r}} \)
Step 2: Substitute values: \( \mathrm{U = (9 \times 10^9) \dfrac{(2.0 \times 10^{-6})(-3.0 \times 10^{-6})}{0.5}} \)
Step 3: \( \mathrm{U = -0.108 \, J} \)
Final Answer: The potential energy of the system is \( \mathrm{-0.108 \, J} \), indicating an attractive interaction.
Example :
Three point charges, \(q_1 = +2.0 \, \mu C\), \(q_2 = +2.0 \, \mu C\), and \(q_3 = -1.0 \, \mu C\), are placed at the vertices of an equilateral triangle of side \(0.3 \, m\). Find the total potential energy of the system.
▶️ Answer/Explanation
Step 1: Formula for total potential energy: \( \mathrm{U_{total} = \dfrac{1}{4 \pi \varepsilon_0} \left( \dfrac{q_1 q_2}{r} + \dfrac{q_1 q_3}{r} + \dfrac{q_2 q_3}{r} \right)} \)
Step 2: Substitute values: \( \mathrm{U_{total} = (9 \times 10^9) \left( \dfrac{(2 \times 10^{-6})(2 \times 10^{-6})}{0.3} + \dfrac{(2 \times 10^{-6})(-1 \times 10^{-6})}{0.3} + \dfrac{(2 \times 10^{-6})(-1 \times 10^{-6})}{0.3} \right)} \)
Step 3: Simplify terms: \( \mathrm{U_{total} = (9 \times 10^9) \dfrac{(4 – 2 – 2) \times 10^{-12}}{0.3}} \)
\( \mathrm{U_{total} = 0} \)
Final Answer: The total potential energy of the three-charge system is \( \mathrm{0 \, J} \). (Repulsive and attractive interactions cancel.)
Example:
A charge \(q = +1.0 \, \mu C\) is moved from point A to point B in the electric field of a point charge \(Q = +4.0 \, \mu C\). If the distance from \(Q\) to A is \(0.4 \, m\) and to B is \(0.2 \, m\), find the work done by the electric field and the change in potential energy of the system.
▶️ Answer/Explanation
Step 1: Potential energy at a distance \(r\): \( \mathrm{U = \dfrac{1}{4 \pi \varepsilon_0} \dfrac{Q q}{r}} \)
Step 2: At A (\(r = 0.4\)): \( \mathrm{U_A = (9 \times 10^9) \dfrac{(4 \times 10^{-6})(1 \times 10^{-6})}{0.4}} \) \( \mathrm{U_A = 0.09 \, J} \)
At B (\(r = 0.2\)): \( \mathrm{U_B = (9 \times 10^9) \dfrac{(4 \times 10^{-6})(1 \times 10^{-6})}{0.2}} \) \( \mathrm{U_B = 0.18 \, J} \)
Step 3: Change in potential energy: \( \mathrm{\Delta U = U_B – U_A = 0.18 – 0.09 = +0.09 \, J} \)
Step 4: Work done by the field: \( \mathrm{W = – \Delta U = -0.09 \, J} \)
Final Answer: \(\Delta U = +0.09 \, J\), \(W = -0.09 \, J\). (The field does negative work since the charge is pushed closer to another positive charge.)