The flat air capacitor was charged to a potential difference of 50 V and disconnected from the source.

The flat air capacitor was charged to a potential difference of 50 V and disconnected from the source. Then a glass plate (ε = 6) was inserted between the plates. Determine the potential difference established in this case.

Given:
U1 = 50V;
e2 = 6;
Find: U2.
Decision:
1) Let’s use the formula to find the electrical capacity.
C = q / U.
Let us express q from it and get: q = CU.
2) we denote for the initial electrical capacity of the capacitor C1, and for the final one – C2. We also take into account that q = const.
Based on this, we get the formula:
C1U1 = C2U2.
3) from the above formula, Let us express the potential difference.
U2 = U1 * C1 / C2.
4) We write down the formula for the electrical capacity of a flat capacitor:
C = e * e0 * S / d.
We can use this formula to find electrical capacities (both initial and final. You can write them in the form of a system.
C1 = e1 * e0 * S / d; C2 = e2 * e0 * S / d.
In these equations, e1 is the dielectric constant of air (it is equal to 1), e2 is the dielectric constant of the glass (it is equal to 6).
5) We solve the system. First, divide the first inequality by the second. And we get:
C1 / C2 = e1 / e2.
Based on this, the formula that we wrote above becomes different:
U2 = U1 * e1 / e2.
6) In order to find the answer, we just need to substitute the values ​​in the formula.
U2 = 50 * 1/6 = 8.33 B.
Answer: 8.33