An initially uncharged air-filled capacitor is connected to a 4.99 V charging source. As a result, the capacitor acquires 3.17×10−5 C of charge. Then, while the capacitor remains connected to the cha

Question

An initially uncharged air-filled capacitor is connected to a 4.99 V charging source. As a result, the capacitor acquires 3.17×10−5 C of charge. 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 𝜅 of this substance is 3.05.

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

When the dielectric is inserted capacitance (C) increases by a factor of 𝜅 (the dielectric constant).

Initially (air-filled capacitor) we have C₀ = Q₀/V₀ = (3.17×10⁻⁵ C)/4.99 V = 6.35 × 10^–6 F.

After inserting the dielectric we have C_f = 𝜅C₀ = 3.05C₀.

C_f = Q_f/V_f = 3.05C₀

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V_f = V₀ = 4.99 V since the capacitor is still connected to the battery (charging source).

Since Q_f = C_fV_f = 3.05C₀V_f = 3.05C₀V₀ and C₀V₀ = Q₀ = 3.17×10⁻⁵ C

then Q_f = 3.05Q₀ = 3.05 × 3.17×10⁻⁵ = 9.67×10⁻⁵ C

Answer

The voltage across the capacitor after the dielectric is inserted is still 4.99 V

The charge stored by the capacitor after the dielectric is inserted increases to 9.67×10⁻⁵ C

[Note: If the battery were disconnected before the dielectric was inserted the charge would stay the same and the voltage across the capacitor would decrease.]