AP Physics C E&M- 10.2 Redistribution of Charge between Conductors- Study Notes- New Syllabus
AP Physics C E&M- 10.2 Redistribution of Charge between Conductors – Study Notes
AP Physics C E&M- 10.2 Redistribution of Charge between Conductors – Study Notes – per latest Syllabus.
Key Concepts:
- Redistribution of Charge Between Conductors
Redistribution of Charge Between Conductors
When two conductors come into physical contact, charges move freely between them until both reach the same electric potential. The redistribution continues until electrostatic equilibrium is established.
Key Principles:
Charge Mobility: Free electrons move between conductors through the contact point.
Equipotential Condition: At equilibrium, both conductors are at the same potential:
\( \mathrm{V_1 = V_2} \)
Total Charge Conservation: The total charge of the system remains constant:
\( \mathrm{Q_{total} = Q_1 + Q_2 = Q’_1 + Q’_2} \)
Final Charge Distribution:
- For identical conductors, charges redistribute equally.
- For different conductors, charges redistribute such that:
\( \mathrm{\dfrac{Q’_1}{R_1} = \dfrac{Q’_2}{R_2}} \)
since potential \( \mathrm{V = \dfrac{1}{4 \pi \varepsilon_0} \dfrac{Q}{R}} \).
Grounding and Ground Potential:
Ground (Earth) is treated as a conductor with an effectively infinite charge reservoir.
The potential of the ground is defined as zero:
\( \mathrm{V_{ground} = 0} \)
If a conductor is connected to ground, charges flow between the conductor and Earth until the conductor’s potential is zero.
Induced Charging by Grounding:
- If an external electric field is applied near a grounded conductor, electrons move between the conductor and Earth.
- This induces a net charge on the conductor that cancels the internal field, maintaining electrostatic equilibrium.
- This process is used in experiments like the Faraday ice pail and in electrostatic shielding.
Physical Consequences:
- Large potential differences between conductors can cause discharge (sparks, static shock).
- Grounding can neutralize an object or induce charge in it depending on the configuration.
Example :
Two identical spheres, one with \( Q_1 = +6.0 \, \mu C \) and the other with \( Q_2 = -2.0 \, \mu C \), are brought into contact and separated. What is the final charge on each?
▶️ Answer/Explanation
Step 1: Total charge: \( \mathrm{Q_{total} = +6.0 + (-2.0) = +4.0 \, \mu C} \).
Step 2: Identical spheres share charge equally: \( \mathrm{Q’_1 = Q’_2 = \dfrac{4.0}{2} = +2.0 \, \mu C} \).
Final Answer: Each sphere has \( +2.0 \, \mu C \) after separation.
Example :
A sphere of radius \( R_1 = 0.05 \, m \) carries \( +3.0 \, \mu C \). It is brought into contact with a sphere of radius \( R_2 = 0.15 \, m \) that is initially uncharged. Find the final charges on each.
▶️ Answer/Explanation
Step 1: Total charge: \( \mathrm{Q_{total} = 3.0 \, \mu C} \).
Step 2: Potentials equalize: \( \mathrm{\dfrac{Q’_1}{R_1} = \dfrac{Q’_2}{R_2}} \).
Step 3: With conservation \( \mathrm{Q’_1 + Q’_2 = 3.0} \), solving gives: \( \mathrm{Q’_1 = 0.75 \, \mu C, \; Q’_2 = 2.25 \, \mu C} \).
Final Answer: Small sphere: \( +0.75 \, \mu C \); Large sphere: \( +2.25 \, \mu C \).
Example:
A neutral conducting sphere is placed near a positively charged rod. The sphere is then grounded while the rod remains nearby. Describe the charge distribution after the ground connection is removed.
▶️ Answer/Explanation
Step 1: The rod polarizes the neutral sphere: electrons shift toward the rod, leaving the far side positively charged.
Step 2: When grounded, electrons flow from Earth into the sphere, neutralizing the positive side.
Step 3: Ground is removed, trapping excess electrons on the sphere.
Final Answer: The sphere acquires a net negative charge due to induced charging by grounding.