Two conducting balls with radii R1 = 1 cm and R2 = 2 cm, respectively, are charged to the potentials F1 = 2 V

Two conducting balls with radii R1 = 1 cm and R2 = 2 cm, respectively, are charged to the potentials F1 = 2 V and F2 = 4 V are connected by a metal wire. What is the potential of the balls connected by a wire?

Let’s translate all the values ​​from given to the SI system:
R1 = 1 cm. = 0.01 m.
R2 = 2 cm = 0.02 m.

Before the balls are connected.
The potentials are equal:
φ1 = q1 / (4π * ε0 * ε * R1).
φ2 = q2 / (4π * ε0 * ε * R2).
Hence the charges are equal:
q1 = 4π * ε0 * ε * R1 * φ1 = 4 * 3.14 * 8.85 * 10 ^ -12 * 1 * 0.01 * 2 = 2.2 * 10 ^ -12 Cl.
q1 = 4π * ε0 * ε * R1 * φ1 = 4 * 3.14 * 8.85 * 10 ^ -12 * 1 * 0.3 * 150 = 8.89 * 10 ^ -12 Cl.
After connecting the balls, their wound potentials between themselves φ1` = φ2`.
φ1` = q1` / (4π * ε0 * ε * R1).
φ2` = q2` / (4π * ε0 * ε * R2).
Let’s equate them:
(q1` / (4π * ε0 * ε * R1)) = (q2` / (4π * ε0 * ε * R2)).
Taking into account that q1` + q2` = q1 + q2, and hence q2` = q1 + q2-q1`:
(q1` / (4π * ε0 * ε * R1)) = ((q1 + q2-q1`) / (4π * ε0 * ε * R2)).
Reduce it by (4π * ε0 * ε), and move everything in one direction:
(q1` / R1) – ((q1 + q2-q1`) / R2 = 0.
Let’s bring to a common denominator:
(q1` * R2 + q1` * R1-R1 * (q1 + q2)) / R1 * R2 = 0.
Let us reduce by R1 * R2 and take q1` out of this expression:
q1` = R1 * (q1 + q2) / (R1 + R2).
Substitute the numbers:
q1` = R1 * (q1 + q2) / (R1 + R2) = 0.01 * ((2.2 * 10 ^ -12) + (8.89 * 10 ^ -12)) / (0.01+ 0.02) = 3.7 * 10 ^ -12 Cl.
Let’s define the potential:
φ1` = q1` / (4π * ε0 * ε * R1) = (3.7 * 10 ^ -12) / (3.14 * 8.85 * 10 ^ -12 * 1 * 0.01) = 3, 3 B.
Answer: φ1` = φ2` = 3.3 V.