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At what distance 'd' from the surface of asolid sphere of radius 'R', (a)potential is sme as at a distance (R)/(2) from the centre ? (b) field strengt is ame as at a distance (R)/(4) fromcentre.

Answer» <html><body><p><br/></p>Solution :(a) Given, `V_("outside") = V_("inside")` <br/> No Such <a href="https://interviewquestions.tuteehub.com/tag/point-1157106" style="font-weight:bold;" target="_blank" title="Click to know more about POINT">POINT</a> will <a href="https://interviewquestions.tuteehub.com/tag/exist-979486" style="font-weight:bold;" target="_blank" title="Click to know more about EXIST">EXIST</a>. Because potential at <a href="https://interviewquestions.tuteehub.com/tag/centre-912170" style="font-weight:bold;" target="_blank" title="Click to know more about CENTRE">CENTRE</a> is `(-1.5GM)/(R)`. Potential at surface is `(-<a href="https://interviewquestions.tuteehub.com/tag/gm-1008640" style="font-weight:bold;" target="_blank" title="Click to know more about GM">GM</a>)/(R)` <br/> and potential at infinity is zero . From centre to surface potential between `(-1.5GM)/(R)` <br/> and `(-GM)/(R)`. From surface to infinity potential varies between `(-GM)/(R)` and zero . <br/> Given, `E_("inside") = E_("outside")` <br/> `:. (GM)/(R^(3))(r_1) = (GM)/(r_(2)^(2))` <br/> Here, `r_(1)` and `r_(2)` are the distances from centre. <br/> `:. (GM)/(R^(3)) ((R)/(4)) = (GM)/((R+d)^(2))` <br/> Solving this <a href="https://interviewquestions.tuteehub.com/tag/equation-974081" style="font-weight:bold;" target="_blank" title="Click to know more about EQUATION">EQUATION</a> , WE get <br/> `d=R`.</body></html>


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