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  4. The electric field between two parallel plates of a capacitor is 2.1 × 10-5 . If a medium is inserted between the plates than the electric field becomes 1.0 × 10-5 . Now, the value of dielectric will be

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Q. The electric field between two parallel plates of a capacitor is $ 2.1\,\times {{10}^{-5}} $ . If a medium is inserted between the plates than the electric field becomes $ 1.0\,\times {{10}^{-5}} $ . Now, the value of dielectric will be

Rajasthan PMTRajasthan PMT 2009

Solution:

$ E=\frac{\sigma }{{{\varepsilon }_{0}}}=2.1\times {{10}^{-5}} $ $ E=\frac{\sigma }{{{\varepsilon }_{0}}K}=1\times {{10}^{-5}} $ $ \therefore $ $ \frac{E}{E}=\frac{\sigma }{{{\varepsilon }_{0}}}\times \frac{{{\varepsilon }_{0}}K}{\sigma }=\frac{2.1\times {{10}^{-5}}}{1\times {{10}^{-5}}} $ or $ K=2.1\approx 2 $