Question

A parallel plate capacitor is filled by a dielectric whose relative permittivity varies with the applied voltage (U) as $ \epsilon = 2U $. A similar capacitor with no dielectric is charged to $ U_0 = 78 \, \text{V} $. It is then connected to the uncharged capacitor with the dielectric. Find the final voltage on the capacitors.

• A. 6V
• B. 8V
• C. 2V
• D. 4V

Hint:

When solving problems involving capacitors with dielectrics, remember that the dielectric affects the capacitance and the voltage. The final voltage in connected capacitors is determined by charge conservation and the dielectric properties of the material.

Answer 1

The Correct Option is A

We are given that the relative permittivity of the dielectric is a function of the applied voltage, \( \epsilon = 2U \). 
Let the initial voltage across the capacitor without the dielectric be \( U_0 = 78 \, \text{V} \). 
The initial energy stored in the capacitor without the dielectric is given by: \[ U_{\text{initial}} = \frac{1}{2} C U_0^2 \] 
Now, when the capacitor with the dielectric is connected to this uncharged capacitor, the voltage across both capacitors will become equal due to charge conservation. 
Let's say the final voltage across both capacitors is \( U_f \). The energy after the connection is: \[ U_{\text{final}} = \frac{1}{2} C \epsilon(U_f) U_f^2 \] Since the charge on both capacitors is conserved, we have: \[ C U_0 = C \epsilon(U_f) U_f \] Substitute the value \( \epsilon(U_f) = 2 U_f \): \[ C U_0 = C \cdot 2 U_f^2 \] \[ U_0 = 2 U_f^2 \] \[ U_f^2 = \frac{U_0}{2} \] \[ U_f = \sqrt{\frac{U_0}{2}} = \sqrt{\frac{78}{2}} = 6 \, \text{V} \] Thus, the final voltage across the capacitors is \( 6 \, \text{V} \).