VAREPSILON }_{0}}A}{d}\] ...(i) where \[{{\varepsilon }_{0}}=\] ELECTRIC PERMITTIVITY of free SPACE K = dielectric constant A = area of each plate of capacitor d = distance between two plates When dielectric (oil) is removed, so capacitance \[{{C}_{0}}=\frac{{{\varepsilon }_{0}}A}{d}\] (ii) Comparing Eqs. (i) and (ii), we GET \[C=K{{C}_{0}}\] \[\Rightarrow \] \[{{C}_{0}}=\frac{C}{K}=\frac{C}{2}(K=2)\]

"> VAREPSILON }_{0}}A}{d}\] ...(i) where \[{{\varepsilon }_{0}}=\] ELECTRIC PERMITTIVITY of free SPACE K = dielectric constant A = area of each plate of capacitor d = distance between two plates When dielectric (oil) is removed, so capacitance \[{{C}_{0}}=\frac{{{\varepsilon }_{0}}A}{d}\] (ii) Comparing Eqs. (i) and (ii), we GET \[C=K{{C}_{0}}\] \[\Rightarrow \] \[{{C}_{0}}=\frac{C}{K}=\frac{C}{2}(K=2)\]

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A parallel plate condenser with oil (dielectric constant 2) between the plates has capacitance C. If oil is removed, the capacitance of capacitor becomes: [AIPMT 1999]

NEET Physics in NEET . 8 months ago

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[d] The capacitance of a parallel plate capacitor with dielectric (oil) between its plates is. \[C=\frac{K{{\VAREPSILON }_{0}}A}{d}\] ...(i) where \[{{\varepsilon }_{0}}=\] ELECTRIC PERMITTIVITY of free SPACE K = dielectric constant A = area of each plate of capacitor d = distance between two plates When dielectric (oil) is removed, so capacitance \[{{C}_{0}}=\frac{{{\varepsilon }_{0}}A}{d}\] (ii) Comparing Eqs. (i) and (ii), we GET \[C=K{{C}_{0}}\] \[\Rightarrow \] \[{{C}_{0}}=\frac{C}{K}=\frac{C}{2}(K=2)\]

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