A $6.0 - \mu \mathrm { F }$ air-filled capacitor is connected across a 100-V potential source (a battery). After the battery fully charges the capacitor, it is left connected and the capacitor is immersed in transformer oil, which has a dielectric constant of 4.5. How much additional charge flows from the battery onto the capacitor during this process?

Question

Quizplus · Accepted Answer

The initial charge on the capacitor is $ Q_0 = C_0 \times V = 6.0 \, \mu \mathrm{F} \times 100 \, \mathrm{V} = 600 \, \mu \mathrm{C} $. When the dielectric is added, the new capacitance is $ C = \kappa \times C_0 = 4.5 \times 6.0 \, \mu \mathrm{F} = 27.0 \, \mu \mathrm{F} $. The new charge is $ Q = C \times V = 27.0 \, \mu \mathrm{F} \times 100 \, \mathrm{V} = 2700 \, \mu \mathrm{C} $. The additional charge is $ Q - Q_0 = 2700 \, \mu \mathrm{C} - 600 \, \mu \mathrm{C} = 2100 \, \mu \mathrm{C} = 2.1 \, \mathrm{mC} $.

A 6.0−μF6.0 - \mu \mathrm { F }6.0−μF Air-Filled Capacitor Is Connected Across a 100-V Potential Source (A

A $6.0 - \mu \mathrm { F }$ Air-Filled Capacitor Is Connected Across a 100-V Potential Source (A