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1.Units, Dimensions and Measurement
medium
The dimensions of permittivity ${\varepsilon _0}$ are
A${A^2}{T^2}{M^{ - 1}}{L^{ - 3}}$
B${A^2}{T^4}{M^{ - 1}}{L^{ - 3}}$
C${A^{ - 2}}{T^{ - 4}}M{L^3}$
D${A^2}{T^{ - 4}}{M^{ - 1}}{L^{ - 3}}$
(AIIMS-2004)
Solution
(b) $F = \frac{1}{{4\pi {\varepsilon _0}}}\,\frac{{{q_1}{q_2}}}{{{r^2}}}$
$ \Rightarrow {\varepsilon _0} = \frac{{|{q_1}|\,|{q_2}|}}{{[F]\,[{r^2}]}} $ $= \frac{{[{A^2}{T^2}]}}{{[ML{T^{ – 2}}]\,[{L^2}]}} $ $= [{A^2}{T^4}{M^{ – 1}}{L^{ – 3}}]$
$ \Rightarrow {\varepsilon _0} = \frac{{|{q_1}|\,|{q_2}|}}{{[F]\,[{r^2}]}} $ $= \frac{{[{A^2}{T^2}]}}{{[ML{T^{ – 2}}]\,[{L^2}]}} $ $= [{A^2}{T^4}{M^{ – 1}}{L^{ – 3}}]$
Standard 11
Physics
