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The force of repulsion between two identical positive charge when kept with a separation r in air is F. Half the gap between the two charges is filled by a dielectric slab of dielectric constant = 4 Then the new force of repulsion between those two charges become |
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Answer» SOLUTION :`F^(1)=1/(4piepsi_(0))(q_(1)q_(2))/((r-t+tsqrtk)^(2))` `F/F^(1)=([r-(r/2)+(r/2)SQRT4]^(2))/r^(2)` = `((3r)/2)^(2)/r^(2)` = `(9/4r^(2))/r^(2)` `F/F^(1)=9/4` `F^(1)=4/9F` |
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