A parallel plate capacitor is filled by a dielectric whose relative permittivity varies with the applied voltage (U) as `epsilon = alpha U` where `alpha = 2V^(-1)`. A similar capacitor with no dielectric is charged to `U_(0) = 78V`. It is then is connected to the uncharged capacitor with the dielectric. Find the final voltage on the capacitors.

Options

A. 2 V
B. 3 V
C. 5 V
D. 6 V

Answer

To solve the problem, we need to analyze the situation step by step.

### Step 1: Identify the parameters
We have two capacitors:
1. Capacitor 1 (C1) is charged to \( U_0 = 78V \) and has no dielectric.
2. Capacitor 2 (C2) has a dielectric whose relative permittivity varies with the voltage as \( \epsilon = \alpha U \), where \( \alpha = 2 \, V^{-1} \).

### Step 2: Calculate the charge on Capacitor 1
The charge \( Q_0 \) on Capacitor 1 when charged to \( U_0 \) is given by:
\[ Q_0 = C_1 U_0 \]

### Step 3: Express the charge on Capacitor 2
The capacitance of Capacitor 2 with the dielectric is given by:
\[ C_2 = \epsilon C = \alpha U C \]
The charge \( Q_2 \) on Capacitor 2 when the voltage across it is \( V \) is:
\[ Q_2 = \epsilon C_2 V = \alpha C V^2 \]

### Step 4: Apply conservation of charge
When the two capacitors are connected, the total charge is conserved. Therefore, we have:
\[ Q_0 = Q_1 + Q_2 \]
Substituting the expressions for the charges:
\[ C_1 U_0 = C_1 V + \alpha C V^2 \]

### Step 5: Rearranging the equation
Rearranging the equation gives:
\[ \alpha C V^2 + C_1 V - C_1 U_0 = 0 \]

### Step 6: Substitute values
Substituting \( \alpha = 2 \, V^{-1} \) and \( U_0 = 78V \):
\[ 2C V^2 + C V - C(78) = 0 \]
Dividing through by \( C \) (assuming \( C \neq 0 \)):
\[ 2V^2 + V - 78 = 0 \]

### Step 7: Solve the quadratic equation
Using the quadratic formula \( V = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 2 \), \( b = 1 \), and \( c = -78 \):
\[ V = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 2 \cdot (-78)}}{2 \cdot 2} \]
\[ V = \frac{-1 \pm \sqrt{1 + 624}}{4} \]
\[ V = \frac{-1 \pm \sqrt{625}}{4} \]
\[ V = \frac{-1 \pm 25}{4} \]

### Step 8: Calculate the possible values for V
Calculating the two possible values:
1. \( V = \frac{24}{4} = 6 \)
2. \( V = \frac{-26}{4} = -6.5 \) (not physically meaningful in this context)

Thus, the final voltage across both capacitors is:
\[ V = 6V \]

### Final Answer:
The final voltage on the capacitors is \( V = 6V \).

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