Question

Suppose two conducting plates are separated by a distance d and held at a potential difference \Delta V by a battery. The capacitance of this configuration, the magnitude of the charge on each plate, and the magnitude of the electric field between the plates are given by C, Q, and E respectively. Next, the battery is carefully disconnected from the plates without letting them touch any other conductors, and sheets of paper are inserted between the plates to completely fill the space. What will be the new values of the following quantities, in terms of the original values ? Q₁ = ?

Answer 1

The charge on the plates Q1 remains the same after the battery is disconnected and a dielectric is inserted between the plates of a capacitor. Other quantities such as capacitance and potential difference are affected, but the charge stays constant.

Step-by-step explanation:

When two conducting plates are separated by a distance d and charged by a battery to voltage Vo, the capacitance C of this configuration and the magnitude of the charge on each plate Q, as well as the electric field E between the plates can be determined. After disconnecting the battery and inserting sheets of paper (dielectric material) between the plates, the quantities we are interested in will be affected.

Since the battery has been disconnected, the charge on the plates remains constant; thus, Q1 = Q. This is because the insertion of the dielectric does not change the total amount of charge, it merely affects the electric field and the potential difference across the capacitor. A dielectric increases the capacitance of the capacitor to C = KoC where K is the dielectric constant of the paper. Consequently, the new potential difference across the plates will be V = Vo/K.

Inserting a dielectric results in a change in the capacitance, potential difference, and electric field, but the charge Q on the plates remains unchanged.