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Find the dimensional formula of epsilon0...

Find the dimensional formula of `epsilon_0` .

Text Solution

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`epsilon_0=Coulomb/newton m^2 `
` epsilon_0=(1)/(KF)xx(q_1q_2)/r^2=M^-1L^-3T^4`
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Knowledge Check

  • The dimensional formula of stress is

    A
    `[ML^(-1) T^(-2)],`
    B
    `[M^@ L^@ T^@]`
    C
    `[MLT^(-1)]`.
    D
    `[MLT^(-2)]`

  • The dimensional formula mu_(0)epsilon_0 is

    A
    `M^0L^(-2)T^2`
    B
    `M^0 L^2T^(-2)`
    C
    `M^(0) L^(1) T^(-1)`
    D
    `M^(0) L^(-1) T^(1)`

  • The dimensional formula of power is

    A
    `ML^(2)T^(-1)`
    B
    `ML^(2)T^(-3)`
    C
    `MLT^(-2)`
    D
    `MLT^(-1)`

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