Question

A parallel-plate capacitor is connected to a battery. A metal sheet of negligible thickness is placed between the plates. The sheet remains parallel to the plates of the capacitor.
(a) The battery will supply more charge.
(b) The capacitance will increase.
(c) The potential difference between the plates will increase.
(d) Equal and opposite charges will appear on the two faces of the metal plate.

Answer 1

Answer:

i hope the answer is option b

Answer 2

The answer is (d), which is opposite and equal charges will appear on the metal plate’s two faces.

Explanation:

• When there is no electric, the electric field between the plates be $\mathrm{E}_{0}=\mathrm{V}_{0} / \mathrm{d}$ between the plates and the potential difference is V0. In the dielectric, the electric field will be  $E=\frac{E_{0}}{K}$

• Then, the potential difference will be:

        $V=\frac{E_{0}}{K} t+E_{0}(d-t)$

where,

d = separation between the capacitor’s plates; t = dielectric slab’s thickness.  

Metal plate is of no thickness, so t=0.  

$V=E_{0} d=V_{0}$

So, the potential stays the same and the charge Q0 also remains constant.

Hence,  

$C=\frac{Q_{0}}{V}=C_{0}$

• On the capacitor, the charge remains constant. So, no charge is supplied to the battery in extra. And here, the metal plate is the dielectric.
• So, on the two faces of the metal plate, equal and opposite charges appear.