Question

A parallel-plate capacitor has capacitance C = 12.5 pF when the volume between the plates is filled with air. The plates are circular, with radius 3.00 cm . The capacitor is connected to a battery and a charge of magnitude 25.0 pC goes onto each plate. With the capacitor still connected to the battery, a slab of dielectric is inserted between the plates, completely filling the space between the plates. After the dielectric has been inserted the charge on each plate has magnitude 45.0 pC . Part A What is the dielectric constant K of the dielectric?

Answer 1

1

Step-by-step explanation:

Initially, the parallel-plate capacitor has a capacitance
`\( C = 12.5 \, \text{pF} \)` and a charge of magnitude
`\( 25.0 \, \text{pC} \)` on each plate. Using the formula
`\( C = (Q)/(V) \)`, where
`\( V \)` is the voltage across the capacitor, the voltage
`\( V \)` is calculated to be
`\( 2.0 \, \text{V} \)`. When a dielectric is inserted, and the charge on each plate becomes
`\( 45.0 \, \text{pC} \)`, the voltage
`\( V' \)` is found to be
`\( 3.6 \, \text{V} \)`.

Using the relationship
`\( C' = K \cdot C \)`, where
`\( C' \)` is the capacitance with the dielectric and
`\( K \)` is the dielectric constant, the capacitance with the dielectric is determined to be
`\( 12.5 * 10^(-12) \, \text{F} \)`. Finally, the dielectric constant
`\( K \)` is calculated by dividing
`\( C' \) by \( C \)`, resulting in a value of 1. Therefore, the dielectric constant of the dielectric material is 1.