Question

Q1.A parallel plate capacitor of capacity C and separation d has a metal plate of thickness t=d/2 placed in between the plates. What is the new capacity?​

Answer 1

Answer:

metal plate of thickness d/2 is placed between the plates, what will be the new capacity ? Solution : As electrons field inside the metal plate is zero, d becomes d/2

Answer 2

Step-by-step explanation:

Given:

Parallel plate capacitor of capacity = C

$Metal \: plate \: thickness =\dfrac{d}{2}$

Plate separation = d

★ To Find:

Capacity of the metal plate.

★ Solution:

$C=\dfrac{\epsilon_0A}{d}$

When a dielectric is filled, thickness t

$C_1=\dfrac{\epsilon_0 A}{d-t+\dfrac{t}{K} }$

$For metal k = infinity, t =\dfrac{d}{2}$

$C_1=\dfrac{2\epsilon_0A}{d}= \boxed{\underline{\underline{2C}}}$

Important points:

When two parallel plates are connected across a battery, the plates are charged and an electric field is established between them, and this setup is known as the parallel plate capacitor.

The parallel plate capacitor formula is given by:

$C=k \epsilon_0 \dfrac{A}{d}$

• k is the relative permittivity of dielectric material.
• d is the separation between the plates.
• A is the area of plates.