Question

The dimension of permittivity of vacuum

• A. $\left[M^{-2}{L}^{-3}{T}^4{A}^2\right]$
• B. $\left[M^{-1}{L}^{-3}{T}^4{A}^3\right]$
• C. $\left[M^{-1}{L}^{-3}{T}^4{A}^2\right]$
• D. $\left[M^{-1}{L}^{-5}{T}^4{A}^2\right]$

Answer 1

Answer: (C)

$\left[M^{-1}{L}^{-3}{T}^4{A}^2\right]$

The permittivity of the vacuum can be given by:

$F_e=\frac{1}{4\pi {\epsilon }_0}\frac{q^2}{r^2}$

$permitivity\left({\epsilon }_0\right)=\frac{(q{)}^2}{\left(F_e\right)\times (r{)}^2}$

$=\frac{[TA{]}^2}{\left[ML{T}^{-2}\right][L{]}^2}$

$=\frac{[TA{]}^2}{\left[M{L}^3{T}^{-2}\right]}$

$=\left[M^{-1}{L}^{-3}{T}^4{A}^2\right]$

$=\left[M^{-1}{L}^{-3}{T}^4{A}^2\right]$

This is the required solution.