Question

A parallel-plate capacitor has a voltage of 391 v applied across its plates, then the voltage source is removed. what is the voltage across its plates if the space between them becomes filled with mica? (the dielectric constant for mica is 5.4 and for air is 1.0006.)

Answer 1

When the space between the plates of a parallel-plate capacitor is filled with a dielectric material, the voltage across the plates decreases. The voltage across the plates when the space between them is filled with mica would be 72.41 V.

Step-by-step explanation:

When the space between the plates of a parallel-plate capacitor is filled with a dielectric material, the voltage across the plates decreases. The relationship between the voltage with air between the plates, Vair, and the voltage with the dielectric, Vmica, is given by:

Vmica = Vair / K

Where K is the dielectric constant of the material. In this case, the dielectric constant for mica is given as 5.4. Therefore, the voltage across the plates when the space between them is filled with mica would be:

Vmica = 391 V / 5.4 = 72.41 V.

Answer 2

When the capacitor is connected to the voltage, a charge Q is stored on its plates. Calling
`C_0` the capacitance of the capacitor in air, the charge Q, the capacitance
`C_0` and the voltage (
`V_0=391 V`) are related by

`C_0 =(Q)/(V_0)` (1)

when the source is disconnected the charge Q remains on the capacitor.

When the space between the plates is filled with mica, the capacitance of the capacitor increases by a factor 5.4 (the permittivity of the mica compared to that of the air):

`C=k C_0 = 5.4 C_0`

this is the new capacitance. Since the charge Q on the plates remains the same, by using eq. (1) we can find the new voltage across the capacitor:

`V=(Q)/(C)=(Q)/(5.4 C_0)`

And since
`Q=C_0 V_0`, substituting into the previous equation, we find:

`V=(C_0 V_0)/(5.4 C_0)=(V_0)/(5.4)=(391 V)/(5.4)=72.4 V`