Question

The capacitance of a capacitor is C when there is air (or vacuum) between its plates. When a dielectric is completely inserted between the plates, the capacitance becomes 2C. Find the dielectric constant of the material.

Hint:

The dielectric constant $K$ strictly indicates the exact multiplying factor by which the total capacitance is enhanced. If it doubles, $K=2$. If it triples, $K=3$. Simple and direct!

Answer 1

Step 1: Understanding the Concept:
Introducing an insulating dielectric material tightly between the metallic plates of a capacitor reduces the effective internal electric field.
Because the potential difference necessarily drops while the charge fundamentally remains the same, the overall capacitance of the system strictly increases.
Step 2: Key Formula or Approach:
The formula strictly relating the original capacitance in a vacuum $C_0$ to the new capacitance $C_d$ with a dielectric is given by $C_d = K C_0$.
Here, $K$ represents the dimensionless dielectric constant (also precisely known as the relative permittivity $\epsilon_r$) of the inserted material.
Step 3: Detailed Explanation:
The initial capacitance with air acting as the medium is securely given as $C_0 = C$.
The final capacitance heavily modified by the complete insertion of the dielectric is explicitly given as $C_d = 2C$.
Substitute these specific measured values completely into the fundamental relationship formula:
\[ 2C = K \times C \] Divide both sides clearly by the non-zero initial capacitance $C$:
\[ K = \frac{2C}{C} \] \[ K = 2 \] Step 4: Final Answer:
The dielectric constant of the inserted material is definitively $2$.