Question

Dimensions of $ {{\varepsilon }_{0}} $ are

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

Answer 1

Answer: (C)

From Coulombs law, $ F=\frac{1}{4\pi {{\varepsilon }_{0}}}\cdot \frac{{{q}_{1}}{{q}_{2}}}{{{r}^{2}}} $ $ \therefore $ $ {{\varepsilon }_{0}}=\frac{1}{4\pi F}\cdot \frac{{{q}_{1}}{{q}_{2}}}{{{r}^{2}}} $ $ \therefore $ Dimension of $ {{\varepsilon }_{0}}=\frac{1}{[ML{{T}^{-2}}]}\times \frac{{{[AT]}^{2}}}{{{[L]}^{2}}} $ $ =[{{M}^{-1}}{{L}^{-3}}{{T}^{4}}{{A}^{2}}] $