Question

A parallel plate capacitor with air between the plates has a capacitance of \(6pF\). What will be the capacitance if the distance between the plates is reduced to half and the space between them is filled with a substance of dielectric constant \(5\)?

• A. \(30pF\)
• B. \(60pF\)
• C. \(15pF\)
• D. \(120pF\)

Hint:

For a capacitor: \[ C \propto \frac{K}{d} \] Reducing distance increases capacitance.

Answer 1

The Correct Option is B

Step 1: Write the capacitance formula.
For a parallel plate capacitor: \[ C=\frac{\varepsilon_0 A}{d} \] When dielectric is introduced: \[ C'=K\frac{\varepsilon_0 A}{d'} \] where: \[ K=5 \]

Step 2: Identify the changes given.
Initial capacitance: \[ C=6pF \] Distance is reduced to half: \[ d'=\frac{d}{2} \] Dielectric constant: \[ K=5 \]

Step 3: Calculate new capacitance.
Since capacitance is inversely proportional to distance: \[ \frac{1}{d'}=\frac{1}{d/2}=\frac{2}{d} \] Thus: \[ C'=K\times 2C \] \[ C'=5\times 2\times 6 \] \[ C'=60pF \]

Step 4: Identify the correct option.
Therefore: \[ \boxed{60pF} \] Hence, the correct answer is: \[ \boxed{\mathrm{(B)}} \]