Question

the capacitor is now reconnected to the battery, and the plate separation is restored to d. a dielectric plate is slowly moved into the capacitor until the entire space between the plates is filled. find the energy u₂ of the dielectric - filled capacitor. the capacitor remains connected to the battery. the dielectric constant is k. express your answer in terms of a, d, v, k, and ε₀. u₂ =

Explanation:

Step1: Recall capacitance formula

The capacitance of a parallel - plate capacitor without dielectric is $C_0=\frac{\epsilon_0A}{d}$. When a dielectric of dielectric constant $\kappa$ is inserted, the new capacitance $C = \kappa C_0=\frac{\kappa\epsilon_0A}{d}$.

Step2: Recall energy formula for capacitor

The energy stored in a capacitor is given by $U=\frac{1}{2}CV^{2}$. Since the capacitor remains connected to the battery, the potential difference $V$ across the capacitor is constant.

Step3: Substitute capacitance into energy formula

Substitute $C=\frac{\kappa\epsilon_0A}{d}$ into $U = \frac{1}{2}CV^{2}$. We get $U_2=\frac{1}{2}\times\frac{\kappa\epsilon_0A}{d}\times V^{2}=\frac{\kappa\epsilon_0AV^{2}}{2d}$.

Answer:

$\frac{\kappa\epsilon_0AV^{2}}{2d}$