Half of the space between parallel plate capa | vedclass

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Question

${\rm{C}} = \frac{{{ \in _0}{\rm{A}}}}{{\rm{d}}}$

${{\rm{C}}_1} = \frac{{{ \in _0}{\rm{KA}}}}{{{\rm{d}}/2}} = \frac{{2{ \in _0}{\rm{KA}}}}{{\rm{d}}

Vedclass · Accepted Answer

$2KC/(1+K)$

Vedclass · Answer

${\rm{C}} = \frac{{{ \in _0}{\rm{A}}}}{{\rm{d}}}$

${{\rm{C}}_1} = \frac{{{ \in _0}{\rm{KA}}}}{{{\rm{d}}/2}} = \frac{{2{ \in _0}{\rm{KA}}}}{{\rm{d}}}$

${{\rm{C}}_2} = \frac{{{ \in _0}{\rm{A}}}}{{{\rm{d}}/2}} = 2\left( {\frac{{{ \in _0}{\rm{A}}}}{{\rm{d}}}} \right)$

$\mathrm{C}_{1} $ and $ \mathrm{C}_{2}$ are in series combination Hence

$\frac{1}{{{C_s}}} = \frac{1}{{{C_1}}} + \frac{1}{{{C_2}}} = \frac{d}{{2{ \in _0}A}}\left( {\frac{1}{K} + \frac{1}{1}} \right)$

$\frac{1}{{{C_s}}} = \frac{d}{{2{ \in _0}A}}\left( {\frac{{K + 1}}{K}} \right)$

${C_s} = \frac{{2{ \in _0}A}}{d}\left( {\frac{K}{{K + 1}}} \right)$

${C_s} = \frac{{2{\rm{CK}}}}{{{\rm{K}} + 1}}$

Half of the space between parallel plate capacitor is filled with a medium of dielectric constant $K$ parallel to the plates . If initially the capacity is $C$, then the new capacity will be

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