Saved Bookmarks
| 1. |
Give relation between vacuum permittivity, vacuum permeability and speed of light. |
|
Answer» Solution :As we know that, `1/(4piepsi_(0))=9XX10^(9)(Nm^(2))/c^(2)andmu_(0)/(4pi)=10^(-7)(Tm)/A` Now, `mu_(0)epsi_(0)=(mu_(0)/(4pi))((4piepsi_(0))/1)` = `(1xx10^(-7))(1/(9xx10^(9)))=1/(9xx10^(16))` = `1/((3xx10^(8))^(2))` but `3xx10^(8)m//s` is the speed of light in VACUUM. `thereforemu_(0)epsi_(0)=1/c^(2)` `thereforec=1/sqrt(mu_(0)epsi_(0))` |
|