Question

A sheet of mica is inserted between the plates of an isolated charged parallel-plate capaci- tor. Mica is a transparent mineral that comes naturally in thin sheets, and is an excellent dielectric. Which of the following statements is true? 1. The capacitance decreases. 2. The charge on the capacitor plates de- creases. 3. The electric field between the capacitor plates increases. 4. The energy of the capacitor does not change. 5. The potential difference across the capac- itor decreases 015 10.0 points In order to store more energy in a parallel- plate capacitor whose plates differ by a fixed voltage. what change would you make in the plates?

Answer 1

Answer:

a) 5.

b) please see below as the answer is self-explanatory.

Explanation:

For a parallel- plate capacitor, applying  Gauss ' Law to a pillbox parallel to the surface of one of the plates, half outside the plate, half inside it, it can be showed that the capacitance can be expressed as follows:

$C = \frac{\epsilon*A}{d} (1)$

So, if we insert a sheet of mica, as it has dielectric constant >1, the value of the capacitance must increase in the same proportion.

By definition, the capacitance obeys the following relationship:

$C = \frac{Q}{V}$

If the capacitance increases due to the insertion of the sheet of mica, and the capacitor is isolated (which means that the charge on the plates must keep the same), the potential difference across the capacitor must decrease, as stated  in the option 5).

b) The energy stored in a capacitor, can be expressed as follows:

$Ue = \frac{1}{2} * C* V^{2}$

If V is fixed, we need to increase the value of C.

If we look to (1), we will find three ways to increase C:

• Increasing the dielectric constant (inserting a sheet of a material of higher dielectric constant)
• Increasing the area of  the plates.
• Decreasing the distance between plates.