Q142 / 150 · MHT CET 2026 (19 April Shift 1) · Physics

A slab of material of dielectric constant $K$ has the same area as the plates of a parallel plate capacitor but has a thickness $(4/5)d$, where $d$ is the separation of the plates. The capacitance in the presence and absence of dielectric are $C$ and $C_0$ respectively. The ratio $(C/C_0)$ is

  • A
    $\dfrac{K + 4}{5K}$
  • B
    $\dfrac{4K}{K + 5}$
  • Correct answer
    C
    $\dfrac{5K}{K + 4}$
  • D
    $\dfrac{K + 5}{4K}$

Solution

The capacitance of a parallel plate capacitor without a dielectric is given by:

$C_0 = \dfrac{\varepsilon_0 A}{d}$

When a dielectric slab of thickness $t$ and dielectric constant $K$ is introduced, the capacitance becomes:

$C = \dfrac{\varepsilon_0 A}{d - t + \dfrac{t}{K}}$

Given $t = \dfrac{4}{5}d$, substituting this value yields:

$C = \dfrac{\varepsilon_0 A}{d - \dfrac{4}{5}d + \dfrac{4d}{5K}}$

$C = \dfrac{\varepsilon_0 A}{\dfrac{d}{5} + \dfrac{4d}{5K}}$

$C = \dfrac{\varepsilon_0 A}{\dfrac{d(K + 4)}{5K}}$

$C = \dfrac{5K \varepsilon_0 A}{d(K + 4)}$

Taking the ratio of $C$ to $C_0$ gives:

$\dfrac{C}{C_0} = \dfrac{5K}{K + 4}$

Answer: $\dfrac{5K}{K + 4}$

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

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