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Define electric flux. Write its SI unit. "Gauss 'law in electrostatic is true for any closed surface, no matter what its shape or size is" Justify this statement with the help of a suitable example. |
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Answer» Solution :For electric flux and its SI UNIT , see To Justify the STATEMENT that Gauss. law is true for any closed SURFACE irrespective of its shape or size ,let us consider two spherical surface of RADII `r_1 and r_2` respectively with a charge q at its centre. Now electric field at the surface of 1ST sphere `E_1 =(1)/( 4 pi in _0).(q)/( r_1^(2)) ` normally outward and flux over this surface. ` phi_1 =int oversetto (E) .oversetto (ds)= E_1 s_1 = ( (1)/( 4 pi in _0) .(q)/( r_1^(2)) ) xx (4 pi r_1^(2)) =(q)/( in_0) ` Again electric field at the surface of 2nd sphere `E_2 =(1)/( 4 pi in _0) .(q)/( r_2^(2))` normallyoutward and flux over this surface. ` "" phi_2 int oversetto (E) . oversetto (ds)=E_2s_2 = ((1)/( 4 pi in _0) .(q)/(r_2^(2))) xx (4 pi r_2^(2)) = (q)/( in_0) ` Thus, it is clear that `phi_1 =phi_2 ` i.e., electric flux is same for both the closed surfaces. 
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