U

Step 1: Understand the situation: We have a capacitor that is filled with a dielectric whose permittivity varies with the applied voltage. The law for the permittivity is given as \( \epsilon = \alpha U \) where \( \alpha = 1 \, \text{V}^{-1} \).
Step 2: Initially, the capacitor with the dielectric is uncharged. The other capacitor (without dielectric) is charged to a voltage \( U_0 = 156 \, \text{V} \).
Step 3: When the charged capacitor is connected in parallel to the uncharged capacitor, charge will redistribute until both capacitors reach a uniform final voltage \( U \).
Step 4: Let's denote the capacitance of the first capacitor with dielectric as \( C_1 \) and that without dielectric as \( C_0 \). Since the first capacitor's permittivity is voltage-dependent, we express the capacitance: \( C_1 = \frac{\epsilon A}{d} = \frac{\alpha U A}{d} = \frac{A}{d} U \) where A is the area of the plates and d is the distance between them.
Step 5: The charge in the charged capacitor is \( Q_0 = C_0 U_0 \). Since both capacitors are in parallel after connection, we have: \( Q_f = Q_0 = C_0 U_0 = C_1 U + C_0 U \).
Step 6: Substitute \( C_1 = \frac{A}{d} U \) into the charge balance equation: \( Q_0 = \frac{A}{d} U + C_0 U \). The total charge in the system can be expressed using the voltage and capacitances.
Step 7: Rearranging gives us a polynomial relation to solve for U: \( C_0 U_0 = \frac{A}{d} U + C_0 U \) leading to \( C_0 U_0 = (C_0 + \frac{A}{d}) U \).
Step 8: Solving for \( U \), we find: \( U = \frac{C_0 U_0}{C_0 + \frac{A}{d}} \).
Step 9: Since we don't have exact values for C_0 and the area A/d, we can't numerically compute \( U \). But the expression gives the final voltage across the capacitors. Thus, the final voltage across both capacitors after connection is \( U \).
Therefore, we conclude that the final voltage \( U \) is a function of the initial charge and the dielectric's varying permittivity.