Capacitance of a parallel plate capacitor becomes 4/3 times its original value if a dielectric slab

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2. Electric Potential and Capacitance

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Capacitance of a parallel plate capacitor becomes $4/3$ times its original value if a dielectric slab of thickness $t = d/2$ is inserted between the plates ($d$ is the separation between the plates). The dielectric constant of the slab is

$8$

$4$

$6$

$2$

Solution

(d) ${C_{air}} = \frac{{{\varepsilon _0}A}}{d}$, with dielectric slab $C'=$ $\frac{{{\varepsilon _0}A}}{{\left( {d – t + \frac{t}{K}} \right)}}$
Given $C' = \frac{4}{3}C \Rightarrow $$\frac{{{\varepsilon _0}A}}{{\left( {d – t + \frac{t}{K}} \right)}} = \frac{4}{3} \times \frac{{{\varepsilon _0}A}}{d}$
$ \Rightarrow K = \frac{{4t}}{{4t – d}} = \frac{{4(d/2)}}{{4[(d/2) – d]}} = 2$

Standard 12

Physics

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Capacitance of a parallel plate capacitor becomes $4/3$ times its original value if a dielectric slab of thickness $t = d/2$ is inserted between the plates ($d$ is the separation between the plates). The dielectric constant of the slab is

Solution

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