Separation between the plates of a parallel plate capacitor is d and the area of each plate is A. When a slab of material of

Q: Separation between the plates of a parallel plate capacitor is $d$ and the area of each plate is $A$. When a slab of material of dielectric constant $k$ and thickness $t(t < d)$ is introduced between the plates, its capacitance becomes

c (c) Potential difference between the plates $V$ = $V_{air}$ + $V_{medium}$ $ = \frac{\sigma }{{{\varepsilon _0}}} \times (d - t) + \frac{\sigma }{{K{\varepsilon _0}}} \times t$ $==>$ $V = \frac{\sigma }{{{\varepsilon _0}}}(d - t + \frac{t}{K})$ $ = \frac{Q}{{A{\varepsilon _0}}}(d - t + \frac{t}{K})$ Hence capacitance $C = \frac{Q}{V} = \frac{Q}{{\frac{Q}{{A{\varepsilon _0}}}(d - t + \frac{t}{K})}}$ $ = \frac{{{\varepsilon _0}A}}{{(d - t + \frac{t}{k})}} = \frac{{{\varepsilon _0}A}}{{d - t\,\left( {1 - \frac{1}{k}} \right)}}$

SECTION - A [PHYSICS - MCQ]

GSEB - English Medium

Separation between the plates of a parallel plate capacitor is $d$ and the area of each plate is $A$. When a slab of material of dielectric constant $k$ and thickness $t(t < d)$ is introduced between the plates, its capacitance becomes

A$\frac{{{\varepsilon _0}A}}{{d + t\left( {1 - \frac{1}{k}} \right)}}$
B$\frac{{{\varepsilon _0}A}}{{d + t\left( {1 + \frac{1}{k}} \right)}}$
C$\frac{{{\varepsilon _0}A}}{{d - t\left( {1 - \frac{1}{k}} \right)}}$
D$\frac{{{\varepsilon _0}A}}{{d - t\left( {1 + \frac{1}{k}} \right)}}$

Diffcult

STD 11 Science
NEET
STD 12 - 2. Electric potential and capacitance
Physics

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