Question

A capacitor has capacitance $C$. When dielectric of constant $3$ completely fills it, new capacitance is:

• A. $C/3$
• B. $C$
• C. $3C$
• D. $9C$

Hint:

Introducing any dielectric medium between the plates of a capacitor always increases its capacitance by a factor of $K$.
Whether the capacitor is connected to a battery or isolated, the capacitance $C'$ is always equal to $K \cdot C$.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
The question asks for the new capacitance of a capacitor when a dielectric material with a dielectric constant of $3$ completely fills the space between its plates.

Step 2: Key Formula or Approach:
The capacitance of a parallel plate capacitor in the presence of a dielectric medium is given by the formula:
\[ C' = \frac{K \epsilon_0 A}{d} = K C \] where $C$ is the initial capacitance in vacuum, $K$ is the dielectric constant of the material, $\epsilon_0$ is the permittivity of free space, $A$ is the area of the plates, and $d$ is the plate separation distance.

Step 3: Detailed Explanation:

• The original capacitance of a parallel plate capacitor without any dielectric (in vacuum or air) is defined by the geometric parameters of the plates and the permittivity of free space. This is mathematically represented as $C = \frac{\epsilon_0 A}{d}$.

• When a dielectric material is introduced, its molecules undergo polarization in the presence of the electric field between the capacitor plates. This polarization creates an internal induced electric field that opposes the applied electric field.

• Due to this opposing induced field, the net electric field ($E$) between the plates is reduced by a factor equal to the dielectric constant ($K$). Thus, the new electric field becomes $E' = \frac{E}{K}$.

• Since the potential difference ($V$) between the plates is directly proportional to the electric field ($V = E \cdot d$), the potential difference also decreases by the same factor, $V' = \frac{V}{K}$.

• Capacitance is defined as the ratio of stored charge to potential difference, $C = \frac{Q}{V}$. When the potential difference decreases to $V'$, the capacity of the system to store charge at a given potential increases.

• Substituting the reduced potential difference $V'$ into the capacitance definition yields the new capacitance: $C' = \frac{Q}{V'} = \frac{Q}{V/K} = K \left(\frac{Q}{V}\right) = K C$.

• In this specific problem, we are given that the dielectric constant $K = 3$ completely fills the space.

• Substituting the value $K = 3$ into our relationship gives the new capacitance as $C' = 3C$.

Step 4: Final Answer:
The new capacitance of the capacitor when completely filled with a dielectric constant of $3$ is $3C$.