Question

An initially uncharged air-filled capacitor is connected to a 5.83 V charging source. As a result, 3.41×10−5 C of charge is transferred from one of the capacitor's plates to the other. Then, while the capacitor remains connected to the charging source, a sheet of dielectric material is inserted between its plates, completely filling the space. The dielectric constant K of this substance is 6.27 . Find the voltage V across the capacitor and the charge Qf stored by it after the dielectric is inserted and the circuit has returned to a steady state.

Answer 1

Q = 23.01 × 10⁻⁵ C

Step-by-step explanation:

given,

capacitor connected to charging source of 5.83 V

Charge (Q) = 3.41 × 10⁻⁵ C

dielectric constant (K) = 6.27

Using formula

Q₀ = C₀ V

3.41 × 10⁻⁵ = C₀ × 5.83

C₀ = 0.584 × 10⁻⁵ F

C(capacitance with dielectric) = K C₀

= 6.27 × 0.584 × 10⁻⁵ F

= 3.67 × 10⁻⁵ F

new charge stored

Q = C V

Q = 3.67 × 10⁻⁵ × 6.27

Q = 23.01 × 10⁻⁵ C