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The dimensions of (1)/(2)epsilon_(0)E^(2) (epsilon_(0)= permitivity of free space and E= electric field) are : |
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Answer» `[ML^(2)T^(-1)]` `:.(1)/(2)in_(0)E^(2)represents=("Energy")/("Volume")` `=(ML^(2)T^(-2))/(L^(3))=[ML^(-1)T^(-2)]` Aliter.`epsilon_(0)=(1)/(4piF).(q_(1)q_(2))/(R^(2))` and `E=(F)/(q)` `:.(1)/(2)epsilon_(0)E^(2)=(1)/(8piF)(q_(1)q_(2))/(r^(2))xx(F^(2))/(q^(2))=(F)/(r^(2))=(MLT^(-2))/(L^(2))` `[ML^(-1)T^(-2)]` Hence the correct choice is `(b)`. |
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