Q132 / 150 · MHT CET 2026 (13 April Shift 2) · Physics

A parallel plate capacitor with air between the plate has a capacitance of $15$ pF. The separation between the plates becomes twice and the space between them is filled with a medium of dielectric constant $3.5$. Then the capacitance becomes $x/4$ pF.
The value of $x$ is

  • Correct answer
    A
    $105$
  • B
    $109$
  • C
    $111$
  • D
    $115$

Solution

The initial capacitance of the parallel plate capacitor with air is given by:

$C_1 = \dfrac{\epsilon_0 A}{d} = 15 \text{ pF}$

When the separation between the plates is doubled ($d' = 2d$) and the space is filled with a dielectric of constant $K = 3.5$, the new capacitance becomes:

$C_2 = \dfrac{K \epsilon_0 A}{d'} = \dfrac{3.5 \epsilon_0 A}{2d}$

Substituting the value of $\dfrac{\epsilon_0 A}{d}$ from the initial condition:

$C_2 = \dfrac{3.5}{2} \times 15 = \dfrac{7}{4} \times 15 = \dfrac{105}{4} \text{ pF}$

Given that the new capacitance is $\dfrac{x}{4} \text{ pF}$, we can compare the two expressions:

$\dfrac{x}{4} = \dfrac{105}{4}$

$x = 105$

Answer: $105$

Hence, the correct answer is option: 'A'.

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