A parallel plate capacitor is filled by a di-electric whose relative permittively varies with the applied voltage according to the law `=alpha V`, where `alpha = 1` per volt. The same (but contiaining no di-electirc) capacitor charged to a voltage `V = 156` volt is connected in parallel to the first ''non-linear'' uncharged capacitor. Determine the final voltage `V_(f)` across the capacitors.

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the Capacitors
We have two capacitors:
1. Capacitor C (charged to 156 V) with no dielectric.
2. Capacitor C' (filled with a dielectric whose relative permittivity varies with voltage as \( K = \alpha V \), where \( \alpha = 1 \, \text{per volt} \)).

### Step 2: Initial Charge on Capacitor C
The charge \( Q \) on capacitor C when charged to voltage \( V = 156 \, \text{V} \) is given by:
\[ Q = C \cdot V = C \cdot 156 \]

### Step 3: Determine the Capacitance of Capacitor C'
The capacitance of capacitor C' with the dielectric is given by:
\[ C' = K \cdot C = \alpha V_f \cdot C \]
where \( V_f \) is the final voltage across both capacitors after they are connected in parallel.

### Step 4: Total Charge Conservation
When the two capacitors are connected in parallel, the total charge is conserved. The charge on capacitor C is transferred to the combined system of C and C':
\[ Q = Q_C + Q_{C'} \]
where \( Q_C = C \cdot V_f \) and \( Q_{C'} = C' \cdot V_f \).

Substituting for \( Q_{C'} \):
\[ Q = C \cdot V_f + C' \cdot V_f = C \cdot V_f + (\alpha V_f \cdot C) \cdot V_f \]
\[ Q = C \cdot V_f + \alpha C \cdot V_f^2 \]

### Step 5: Substitute for Q
Substituting \( Q = C \cdot 156 \) into the charge conservation equation:
\[ C \cdot 156 = C \cdot V_f + \alpha C \cdot V_f^2 \]
Since \( \alpha = 1 \):
\[ C \cdot 156 = C \cdot V_f + C \cdot V_f^2 \]

### Step 6: Simplify the Equation
Dividing through by \( C \) (assuming \( C \neq 0 \)):
\[ 156 = V_f + V_f^2 \]
Rearranging gives:
\[ V_f^2 + V_f - 156 = 0 \]

### Step 7: Solve the Quadratic Equation
Using the quadratic formula \( V_f = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 1, b = 1, c = -156 \):
\[ V_f = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot (-156)}}{2 \cdot 1} \]
\[ V_f = \frac{-1 \pm \sqrt{1 + 624}}{2} \]
\[ V_f = \frac{-1 \pm \sqrt{625}}{2} \]
\[ V_f = \frac{-1 \pm 25}{2} \]
Calculating the two possible values:
1. \( V_f = \frac{24}{2} = 12 \, \text{V} \)
2. \( V_f = \frac{-26}{2} = -13 \, \text{V} \) (not physically meaningful)

Thus, the final voltage across the capacitors is:
\[ V_f = 12 \, \text{V} \]

### Final Answer:
The final voltage \( V_f \) across the capacitors is \( 12 \, \text{V} \).

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