Two spherical conductors are separated by a large distance. They each carry the same positive charge [Q]. Conductor [A] has a larger radius than conductor [B].

Compare the potential at the surface of conductor [A] with the potential at the surface of conductor [B].

• [V_A < V_B]
  • [V_B] is equal to the inverse of the scaler difference between the two sperical radii.

The two conductors are now connected by a wire. How do the potentials at the conductor surfaces compare now?

• [V_A = V_B]
  • By connecting the charges through a wire it causes the potential charge to equalize.

What happens to the charge on conductor [A] after it is connected to conductor [B] by the wire?

• [Q_A] increases
  • Since sphere [A] has a greater radius this mean that it has a lower charge potential and in turn when the charges potentials equalize when the wire is attached the sphere with large radius [A] with have the scaler multiple of the difference between their radii as it's new charge.

Two parallel plates of equal area carry equal and opposite charge [Q_0]. The potential difference between the two plates is measured to be [V_0]. An uncharged conducting plate (the green thing in the picture below) is slipped into the space between the plates without touching either one. The charge on the plates is adjusted to a new value [Q_1] such that the potential difference between the plates remains the same.

Compare [Q_1] and [Q_0]

• [Q_1 = Q_2]
  • Introducing an conductor does not change the charge of the system.

Compare the capacitance of the two configurations in the above problem.

• [C_1 > C_0]
  • Since [C \equiv \frac{ Q}{ \Delta V}] and the potential difference is lower in the instance with the conducting plate present it follows that the capacitance of the according instance will be higher.