Capacitor With Dielectrics: Practice Problem & Solution

Solution Explained:

To solve this problem, we apply the core principles of Capacitor With Dielectrics. Understanding the underlying formula is key to arriving at the correct answer below:

Let’s solve the problem step by step:

Given:

• Initial capacitance:
  $C$ 
• The separation between the plates is doubled, and a dielectric medium is inserted.
• The new capacitance becomes
  $2C$ 

  .

Step 1: Capacitance formula

The capacitance of a parallel plate capacitor is given by:

$$C = \frac{\varepsilon_0 A}{d}$$

where:

• 
  $C$ 

  is the capacitance,

• 
  $\varepsilon_0$ 

  is the permittivity of free space,

• 
  $A$ 

  is the area of the plates, and

• 
  $d$ 

  is the separation between the plates.

Step 2: Effect of doubling the separation and adding a dielectric

When the separation

$d$

is doubled, the capacitance would normally decrease by a factor of 2 (since capacitance is inversely proportional to

$d$

).

Now, when a dielectric of dielectric constant

$K$

is inserted, the capacitance increases by a factor of

$K$

. So, the new capacitance

$C'$

is:

$$C' = K \cdot \frac{\varepsilon_0 A}{2d} = K \cdot \frac{C}{2}$$

We are told that the new capacitance is

$2C$

, so:

$$K \cdot \frac{C}{2} = 2C$$

Step 3: Solve for $K$

$$K = 4$$

Final Answer:

The dielectric constant of the medium is 4.