MP Board · Class 12 · Physics · Electrostatic Potential and CapacitanceA parallel plate capacitor with air between the plates has a capacitance of $8 \text{ pF}$. What will be the capacitance if the distance between the plates is reduced by half, and the space between them is filled with a substance of dielectric constant $K = 6$?

Step-by-Step Solution:

1. Given Data:

   • Initial capacitance with air, $C = 8 \text{ pF} = 8 \times 10^{-12} \text{ F}$
   • Dielectric constant of the substance, $K = 6$

2. Formula for Parallel Plate Capacitor with Air: The capacitance of a parallel plate capacitor with air medium is given by: $$C = \frac{\varepsilon_0 A}{d}$| where $A$ is the area of the plates and $d$ is the initial separation between the plates.

3. Formula for Capacitor with Dielectric: When the space between the plates is filled with a dielectric of constant $K$, the new capacitance $C'$ is given by: $$C' = \frac{K \varepsilon_0 A}{d'}$$ where $d'$ is the new separation between the plates.

4. Applying the Given Condition: According to the problem, the distance is reduced by half: $$d' = \frac{d}{2}$$

5. Substituting $d'$ into the New Capacitance Equation: $$C' = \frac{K \varepsilon_0 A}{(d / 2)}$$ $$C' = \frac{2 K \varepsilon_0 A}{d}$|

6. Relating $C'$ to the Initial Capacitance $C$: Since $C = \frac{\varepsilon_0 A}{d}$, we can substitute this into the expression for $C'$: $$C' = 2 \cdot K \cdot C$$

7. Calculating the Final Value: Substitute the given values of $K$ and $C$: $$C' = 2 \times 6 \times (8 \text{ pF})$$ $$C' = 12 \times 8 \text{ pF}$$ $$C' = 96 \text{ pF}$|

Final Answer:\nThe new capacitance of the capacitor is $96 \text{ pF}$ (or $9.6 \times 10^{-11} \text{ F}$).