Question

An air filled parallel plate capacitor has a capacity \(2\ \text{pF}\). The separation of the plates is doubled and the interspace between the plates is filled with dielectric material, then the capacity is increased to \(6\ \text{pF}\). The dielectric constant of the material is

• A. 3
• B. 6
• C. 2
• D. 4

Hint:

When separation changes, first find the capacitance without the dielectric: doubling \(d\) alone would halve the capacitance. Then multiply by \(K\) to get the final value. Set up the ratio \(\frac{C_2}{C_1} = \frac{K}{2}\) to solve quickly.

Answer 1

The Correct Option is B

Step 1: Read the changes.
An air capacitor has $2$ pF. Then we double the gap between the plates and fill it with a dielectric, and the value rises to $6$ pF. We want the dielectric constant $K$.

Step 2: Recall the capacitor formula.
With air, $C = \dfrac{\varepsilon_0 A}{d}$. With a dielectric of constant $K$ and a new gap, $C' = \dfrac{K \varepsilon_0 A}{d'}$.

Step 3: Note the new gap.
The gap is doubled, so $d' = 2d$. That gives $C' = \dfrac{K \varepsilon_0 A}{2d}$.

Step 4: Tie it to the old value.
Notice $\dfrac{\varepsilon_0 A}{d}$ is just the old capacitance $C_1 = 2$ pF, so $C' = \dfrac{K}{2} C_1$.

Step 5: Plug in the numbers.
$6 = \dfrac{K}{2} \times 2 = K$.

Step 6: State the answer.
So $K = 6$, which is option (B). Doubling the gap halves the capacitance, and the dielectric of $6$ brings it back up sixfold, netting a threefold rise from $2$ to $6$ pF.
\[ \boxed{K = 6} \]