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The electric force between two electric ...

The electric force between two electric charges is given by `F= (1)/(4pi epsi_(0))(q_(1)q_(2))/(r^(2))`. where is a distance . between charges `q_(1) and q_(2)` . So, unit and dimensional formula of `epsi_(0)`, is ...... and...... respectively.

Text Solution

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Here electric force `F=-(1)/(4pi epsi_(0))(q_(1)q_(2))/(r^(2))` is given.
`:.epsi_(0)=(1)/(4piF)(q_(1)q_(2))/(r^(2))`
Putting units of f= newton `q_(1) and q_(2)=` coulomb and r = m in above equation, we get
unit of `epsi_(0)` is `(c*c)/(m*m^(2))=N^(-1)C^(2)m^(-2)`
Now putting dimensional formula of
`-[F]=M^(1)L^(1)T^(-2)`
`[q_(1)]=A^(1)T^(1)`
`[r]=L^(1)`
`:.[ epsi_(0)]=(1)/(M^(1)L^(1)T^(-2))=(A^(2)T^(2))/(L^(2))=M^(-1)L^(-3)T^(4)A^(2)`
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The electric force between two electric charges is given by F=(1)/(4pi epsi_(0))(q_(1)q_(2))/(r^(2)) , where r is the distance between q_(1) and q_(2) . Give the unit and dimensional formula of epsi_(0) is....

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Knowledge Check

  • According to Coulomb's law force between two charges q_(1) and q_(2) at 'r' distance apart....

    A
    `F prop (q_(1)q_(2))/(r)`
    B
    `F prop (q_(1)q_(2))/(r^(2))`
    C
    `F prop ((q_(1)q_(2))/(r))^(2)`
    D
    `F prop (q_(1)+q_(2))/(r^(2))`

  • If electric charge on capacitor formed by two plates of area A is Q, then electric field between them is …..

    A
    `E=(Q)/(epsilon_(0)A)`
    B
    `E=(Q)/(2epsilon_(0)A)`
    C
    `E=(2Q)/(epsilon_(0)A)`
    D
    `(epsilon_(0)A)/(Q)`

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