A parallel-plate capacitor, with air between the plates, is connected across a v

Question

A parallel-plate capacitor, with air between the plates, is connected across a voltage source. This source establishes a potential difference between the plates by placing charge of magnitude4.07 ×10 -6 C on each plate. The space between the plates is then filled with a dielectric material, with a dielectric constant of 7.74. What must the magnitude of the charge on each capacitor plate now be, to produce the same potential difference between the plates as before?
A parallel-plate capacitor, with air between the plates, is connected across a voltage source. This source establishes a potential difference between the plates by placing charge of magnitude4.07 ×10 -6 C on each plate. The space between the plates is then filled with a dielectric material, with a dielectric constant of 7.74. What must the magnitude of the charge on each capacitor plate now be, to produce the same potential difference between the plates as before?
A parallel-plate capacitor, with air between the plates, is connected across a voltage source. This source establishes a potential difference between the plates by placing charge of magnitude4.07 ×10 -6 C on each plate. The space between the plates is then filled with a dielectric material, with a dielectric constant of 7.74. What must the magnitude of the charge on each capacitor plate now be, to produce the same potential difference between the plates as before?

Explanation / Answer

the relation between capacitor,charge and voltage

C = Q / V

Initial Q = 4.07 × 10^-6 C

final Q = (7.74)(4.07 × 10^-6 C)

= 3.15*10^-5 C

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