Email
Chat with tutors
logo

Ask Questions, Get Answers

X
 
Answer
Comment
Share
Q)

Let $ \varepsilon_0$ denote the dimensional formula of the permittivity of vacuum. If M = mass, L = length and T = time and A = electric current then,

$\begin {array} {1 1} (A)\;[ \varepsilon_0 ]=[ M^{-1}L^{-3}T^4A^2] & \quad (B)\;[ \varepsilon_0 ]=[ M^{-1}L^{2}T^{-1}A^{-2}] \\ (C)\;[ \varepsilon_0 ]=[ M^{-1}L^{2}T^{-1}A] & \quad (D)\;[ \varepsilon_0 ]=[ M^{-1}L^{-3}T^2A ] \end {array}$

 

1 Answer

Home Ask Tuition Questions
Your payment for is successful.
Continue
...