Question

The capacitance of a capacitor with dielectric as a medium is $60\mu \text{F}$. When dielectric is removed, its capacitance changes to $12\mu \text{F}$. The permittivity of the dielectric medium is -

• A. $4.4\times {10}^{–11}{\text{C}}^2{\text{N}}^{–1}{\text{m}}^{–2}$
• B. $5\times {10}^{–11}{\text{C}}^2{\text{N}}^{–1}{\text{m}}^{–2}$
• C. $5.00{\text{C}}^2{\text{N}}^{–1}{\text{m}}^{–2}$
• D. $1.85\times {10}^{–11}{\text{C}}^2{\text{N}}^{–1}{\text{m}}^{–2}$

Answer 1

The correct option is A $4.4\times {10}^{–11}{\text{C}}^2{\text{N}}^{–1}{\text{m}}^{–2}$

Dielectric constant of medium,

$K=\frac{C_{\text{dielectric}}}{C_{\text{air}}}=\frac{60}{12}=5$

Permittivity of medium,

$ϵ=K{ϵ}_0=5\times 8.85\times {10}^{-12}$

$=4.4\times {10}^{–11}{\text{C}}^2{\text{N}}^{–1}{\text{m}}^{–2}$