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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 is a distance . between charges q_(1) and q_(2) . So, unit and dimensional formula of epsi_(0), is ...... and...... respectively.

Answer» <html><body><p>`N^(1)<a href="https://interviewquestions.tuteehub.com/tag/c-7168" style="font-weight:bold;" target="_blank" title="Click to know more about C">C</a>^(2)m^(-2),M^(-1)L^(-<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>)T^(4)A^(2)`<br/>`N^(-1)<a href="https://interviewquestions.tuteehub.com/tag/cm-919986" style="font-weight:bold;" target="_blank" title="Click to know more about CM">CM</a>^(-2),M^(-1)L^(-1)T^(-1)A`<br/>`N^(2)C^(2)m^(-2)M^(-1)L^(-3)T^(4)A^(2)`<br/>`NCm^(-2),MLTA`<br/></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :`F=(1)/(<a href="https://interviewquestions.tuteehub.com/tag/4pi-1882352" style="font-weight:bold;" target="_blank" title="Click to know more about 4PI">4PI</a> epsi_(0)).(q_(1)q_(2))/(r^(2))` <br/> `:.epsi_(0)=(q_(1)q_(2))/(4pi Fr^(2))` <br/> `:.` Units of `epsi_(0)=(C^(2))/(Nm^(2))=N^(-1)C^(2)m^(-2)` <br/> `[epsi_(0)]=([C^(2)])/([N][m^(2)])=(A^(2)T^(2))/((M^(1)L^(1)T^(-2))(L^(2)))` <br/> `=M^(-1)L^(-3)T^(4)A^(2)`</body></html>


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