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 magnitude 4.33 * 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?

Answer: 33.5 x 10⁻⁶ C

Explanation:

By definition, the capacitance is the relationship between the charge on one of the plates (assuming it's a capacitor) and the voltage between them:

C = Q / V

Now, it can be showed that we can find the capacitance of a parallel-plate capacitor, taking into account the geometry and the dielectric material only, as follows:

C = ε A / d = ε₀ εr A / d

If the dielectric is air, we can assume εr = 1

If the space between plates is filled with a dielectric of dielectric constant 7.74, the new value for capacitance (regarding the former value) must be 7.74 times larger, as A and d didn't change.

So, in order to produce the same potential difference between the plates, we need to increase the charge, exactly 7.7 times:

Q = 7.7. 4.33. 10⁻⁶ C = 33.5. 10⁻⁶ C