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  4. The plates in a parallel plate capacitor are separated by a distance d with air as the medium between the plates. In order to increase the capacity by 66% a dielectric slab of dielectric constant 5 is introduced between the plates. What is the thickness of the dielectric slab?

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Q. The plates in a parallel plate capacitor are separated by a distance d with air as the medium between the plates. In order to increase the capacity by 66% a dielectric slab of dielectric constant 5 is introduced between the plates. What is the thickness of the dielectric slab?

EAMCETEAMCET 2010Electrostatic Potential and Capacitance

Solution:

The capacity in air $ C=\frac{{{\varepsilon }_{0}}A}{d} $ The capacity when dielectric slab of dielectric constant 5 is introduced between the plates $ C=\frac{{{\varepsilon }_{0}}A}{d-t+\frac{t}{5}} $ $ \therefore $ $ \frac{C}{C}=\frac{d-t+\frac{t}{5}}{d} $ $ C=\frac{166}{100}C $ $ \frac{100}{166}=\frac{d-t+\frac{t}{5}}{d}=\frac{d-\frac{4t}{5}}{d} $ or $ 100d=166d-166\left( \frac{4t}{5} \right) $ $ 166\left( \frac{4t}{5} \right)=66d $ $ t=\frac{66d\times 5}{166\times 4}=\frac{d}{2} $