Question

Find the dimensions of electric permittivity

• A. $ [{{A}^{2}}{{M}^{-1}}{{L}^{-3}}{{T}^{4}}] $
• B. $ [{{A}^{2}}{{M}^{-1}}{{L}^{3}}{{T}^{0}}] $
• C. $ [A{{M}^{-1}}{{L}^{-3}}{{T}^{4}}] $
• D. $ [{{A}^{2}}{{M}^{0}}{{L}^{-3}}{{T}^{4}}] $

Answer 1

Answer: (A)

From Coulombs law, the force of attraction/repulsion between two point charges q, q separated by distance $ r $ is $ F=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{{{q}^{2}}}{{{r}^{2}}} $ $ \Rightarrow $ $ {{\varepsilon }_{0}}=\frac{1}{4\pi }.\frac{{{q}^{2}}}{F{{r}^{2}}} $ where $ {{\varepsilon }_{0}} $ is electric permittivity. Dimensions of $ {{\varepsilon }_{0}}=\frac{{{[AT]}^{2}}}{[ML{{T}^{-2}}][{{L}^{2}}]} $ $ {{\varepsilon }_{0}}=[{{A}^{2}}{{M}^{-1}}{{L}^{-3}}{{T}^{4}}] $ Note: The constant $ {{\varepsilon }_{0}} $ (epsilon zero) is called permittivity of free space and its vaiue is $ 8.85\times {{10}^{-12}} $ $ Couloum{{b}^{2}}/newton\text{ }metr{{e}^{2}} $ .