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$\frac{1}{2} \varepsilon_0 E ^2$ નું પારિમાણિક સૂત્ર શું થાય?
જ્યાં $\varepsilon_0$ મુક્ત અવકાશની પરમિટિવિટી અને $E$ વિદ્યુતક્ષેત્ર છે.
A$M^1L^2T^{-2}$
B$M^1L^{-1}T^{-2}$
C$M^1L^2T^{-1}$
D$MLT^{-1}$
(AIPMT-2010) (AIIMS-2014) (IIT-2000)
Solution
Energy, density, of, an ,electric, filed, E, is
${u_E} = \frac{1}{2}{\varepsilon _o}{E^2}$
Where,${\varepsilon _o}$, is, permittivity, of, free, space
${u_E} = \frac{{Energy}}{{Volume}} = \frac{{M{L^2}{T^{ – 2}}}}{{{L^3}}} = M{L^{ – 1}}{T^{ – 2}}$
Hence, the, dim ension, of, $\frac{1}{2}{\varepsilon _o}{E^2}\,is\,M{L^{ – 1}}{T^{ – 2}}$
${u_E} = \frac{1}{2}{\varepsilon _o}{E^2}$
Where,${\varepsilon _o}$, is, permittivity, of, free, space
${u_E} = \frac{{Energy}}{{Volume}} = \frac{{M{L^2}{T^{ – 2}}}}{{{L^3}}} = M{L^{ – 1}}{T^{ – 2}}$
Hence, the, dim ension, of, $\frac{1}{2}{\varepsilon _o}{E^2}\,is\,M{L^{ – 1}}{T^{ – 2}}$
Standard 11
Physics
