AP Physics C E&M- 9.3 Conservation of Electric Energy- Study Notes- New Syllabus
AP Physics C E&M- 9.3 Conservation of Electric Energy – Study Notes
AP Physics C E&M- 9.3 Conservation of Electric Energy – Study Notes – per latest Syllabus.
Key Concepts:
- Conservation of Electric Energy
Conservation of Electric Energy
A charged particle in an electric field experiences changes in its electric potential energy and kinetic energy. The total mechanical energy of the system is conserved, provided only electrostatic forces are acting (no non-conservative forces like friction).
Work–Energy Relation:
The work done by the electric field when moving a charge \(q\) between two points A and B is equal to the negative change in electric potential energy:
\( \mathrm{W_{field} = – \Delta U = U_A – U_B} \)
Potential Difference and Energy Change:
The change in potential energy is related to the electric potential difference (\(\Delta V\)) between two points:
\( \mathrm{\Delta U = q \Delta V} \)
- If \(\mathrm{q > 0}\): the charge moves spontaneously from higher to lower potential.
- If \(\mathrm{q < 0}\): the charge moves spontaneously from lower to higher potential.
Conservation of Mechanical Energy:
As the electric field does work, potential energy converts into kinetic energy or vice versa:
\( \mathrm{\Delta K + \Delta U = 0 \quad \Rightarrow \quad K_A + U_A = K_B + U_B} \)
Integral Form:
The potential difference can also be expressed using the electric field:
\( \mathrm{\Delta V = – \displaystyle \int_A^B \vec{E} \cdot d\vec{l}} \)
Applications:
- Motion of charges in capacitors and parallel plate fields.
- Energy conversion in circuits and batteries.
- Particle accelerators (charges gain kinetic energy when accelerated through a potential difference).
Example :
An electron (\( q = -1.6 \times 10^{-19} \, C \)) is accelerated from rest through a potential difference of \(1000 \, V\). Find its final speed (ignore relativistic effects).
▶️ Answer/Explanation
Step 1: Energy gained by electron: \( \mathrm{\Delta K = – \Delta U = q \Delta V = (1.6 \times 10^{-19})(1000)} \)
\( \mathrm{\Delta K = 1.6 \times 10^{-16} \, J} \)
Step 2: Kinetic energy: \( \mathrm{\tfrac{1}{2} m v^2 = \Delta K} \)
\( \mathrm{v = \sqrt{\tfrac{2 \Delta K}{m}} = \sqrt{\tfrac{2 (1.6 \times 10^{-16})}{9.11 \times 10^{-31}}}} \)
Step 3: \( \mathrm{v \approx 1.87 \times 10^7 \, m/s} \)
Final Answer: The electron’s final speed is approximately \( \mathrm{1.9 \times 10^7 \, m/s} \).
Example :
A proton is released from rest at a location with potential \(V_A = 120 \, V\). It moves to a point with potential \(V_B = 30 \, V\). Find the change in potential energy and the proton’s final kinetic energy.
▶️ Answer/Explanation
Step 1: Potential difference: \( \mathrm{\Delta V = V_B – V_A = 30 – 120 = -90 \, V} \)
Step 2: Change in potential energy: \( \mathrm{\Delta U = q \Delta V = (1.6 \times 10^{-19})(-90)} = -1.44 \times 10^{-17} \, J \)
Step 3: Since total energy is conserved: \( \mathrm{\Delta K = – \Delta U = +1.44 \times 10^{-17} \, J} \)
Final Answer: Potential energy decreases by \( \mathrm{1.44 \times 10^{-17} \, J} \), and the proton gains the same amount of kinetic energy.