A parallel-plate capacitor of area A,  plate separation d and capacitance C is filled wih four dielectric materials having dielectric constants k1, k2, k3 and k4 as shown in the figure below. If a single dielectric material is to be used to have the same capacitance C in this capacitor, then its dielectric constant k is given by

  • Option 1)

    k = k_1 + k_2 + k_3 + 3k_4

  • Option 2)

    k = \frac{2} {3}\left( {k_1 + k_2 + k_3 } \right) + 2k_4

  • Option 3)

    \frac{2} {k} = \frac{3} {{k_1 + k_2 + k_3 }} + \frac{1} {{k_4 }}

  • Option 4)

    \frac{1} {k} = \frac{1} {{k_1 }} + \frac{1} {{k_2 }} + \frac{1} {{k_3 }} + \frac{3} {{2k_4 }}

 

Answers (1)

As we discussed

If K filled between the plates -

{C}'=K\frac{\epsilon _{0}A}{d}={C}'=Ck

 

 

- wherein

C\propto A

C\propto\frac{1}{d}

 

 and

Parallel Grouping -

C_{eq}=C_{1}+C_{2}+cdots

- wherein

 

 C_{1}=\frac{2\varepsilon _{oK_{1A}}}{3d},C_{2}=\frac{2K_{2}\varepsilon oA}{3d}

C_{3}=\frac{2\varepsilon _{oK_{3}A}}{3d},C_{4}=\frac{_{2}\varepsilon oK_{4}A}{d}

\frac{1}{C_{AB}}=\frac{1}{C_{1}+C_{2}+C_{3}}+\frac{1}{C_{4}}=\frac{1}{K(\frac{\epsilon oA}{d})}=\frac{1}{\frac{2\epsilon oA}{3d}}(K_{1}+K_{2}+K_{3})+\frac{1}{\frac{2\epsilon oA}{d}}K_{4}

\frac{1}{K}=\frac{3}{2(K_{1+K_{2}+K_{3}})}+\frac{1}{2K_{4}}\\ \frac{2}{K}=\frac{3}{(K_{1+K_{2}+K_{3}})}+\frac{1}{2K_{4}}=\frac{2}{K}=\frac{3}{(K_{1+K_{2}+K_{3}})}+\frac{1}{K_{4}}


Option 1)

k = k_1 + k_2 + k_3 + 3k_4

This option is incorrect 

Option 2)

k = \frac{2} {3}\left( {k_1 + k_2 + k_3 } \right) + 2k_4

This option is incorrect 

Option 3)

\frac{2} {k} = \frac{3} {{k_1 + k_2 + k_3 }} + \frac{1} {{k_4 }}

This option is correct 

Option 4)

\frac{1} {k} = \frac{1} {{k_1 }} + \frac{1} {{k_2 }} + \frac{1} {{k_3 }} + \frac{3} {{2k_4 }}

This option is incorrect 

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