Question

The capacitance of a parallel plate capacitor with air as medium is 3 $\mu F$. With the introduction of a dielectric medium between the plates, the capacitance becomes 15 $\mu F$. The permittivity of the medium is:

• A. $5C^2{N}^{-1}{m}^{-2}$
• B. $15C^2{N}^{-1}{m}^{-2}$
• C. $0.44\times {10}^{-10}{C}^2{N}^{-1}{m}^{-2}$
• D. $0.41\times {10}^{-10}{C}^2{N}^{-1}{m}^{-2}$

Answer 1

The correct option is C $0.44\times {10}^{-10}{C}^2{N}^{-1}{m}^{-2}$

Capacitance of a paralle plate capacitor $C=\frac{\epsilon A}{d}$

here $\epsilon$ is the permitivity of the medium

For air medium $\epsilon ={\epsilon }_0=8.85\times {10}^{-12}{C}^2{N}^{-1}{m}^{-2}\therefore C^{\prime }=\frac{{\epsilon }_{0}A}{d}....1$

whearas for dielectric medium capacitance $C=\frac{\epsilon A}{d}...2$

From $1and2$

$\frac{C}{C^{\prime }}=\frac{\epsilon }{{\epsilon }_0}$

$\therefore \epsilon =\frac{15}{3}\times 8.85\times {10}^{-12}=(44.25\times {10}^{-12})=0.44\times {10}^{-10}{C}^2{N}^{-1}{m}^2$

Hence the $permitivityofthemedium\epsilon =0.44\times {10}^{-10}{C}^2{N}^{-1}{m}^2$