A parallel plate capacitor of capacitance C0 has plates of area A with separation d between them.When it is connected to a battery of voltage V0, it has charge of magnitude Q0 on its plates.It is then disconnected from the battery and the space between the plates is filled with a material of dielectric constant 3.After the dielectric is added, the magnitudes of the charge on the plates and the potential difference between them are:
A)Q0, V0.
B)Q0, V0.
C)Q0, V0.
D)Q0, 3V0.
E)3Q0, 3V0.

Question

A parallel plate capacitor of capacitance C0 has plates of area A with separation d between them.When it is connected to a battery of voltage V0, it has charge of magnitude Q0 on its plates.It is then disconnected from the battery and the space between the plates is filled with a material of dielectric constant 3.After the dielectric is added, the magnitudes of the charge on the plates and the potential difference between them are:
A)Q0, V0. 
B)Q0, V0. 
C)Q0, V0. 
D)Q0, 3V0. 
E)3Q0, 3V0.

Quizplus · Accepted Answer

When the dielectric is added, the capacitance of the capacitor increases. Capacitance C is given by:
C = ε0 * k * A / d,
where ε0 is the permittivity of free space, k is the dielectric constant, A is the area of the plates, and d is the separation between them.
Since the dielectric constant k increases by a factor of 3, the capacitance C increases by a factor of 3. Therefore, the charge Q on the plates remains the same, but the potential difference V between the plates decreases by a factor of 3. So, the magnitudes of the charge on the plates and the potential difference between them are Q0 and V0/3, respectively.

A Parallel Plate Capacitor of Capacitance C0 Has Plates of Area