InterviewSolution
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InterviewSolution
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मान गोलाथ सम्मित आवेश वितरण में आवेश घनत्व r=R तक `rho(r)=rho_(0)(5/4-(r )/(R ))` के अनुसार बदलता है तथा `rgtR` तक `rho(r )=0` बदलता है जहाँ r मूल बिंदु से दुरी है मूल बिंदु से `r(rltR)` दुरी पर विधुत क्षेत्र हैA. `(rho_(0)r)/(3epsilon_(0))(5/4)-(r/R)`B. `(4pi rho_(0)r)/(3epsilon_(0))(5/3)-(r/R)`C. `(rho_(0)r)/(4epsilon_(0))(5/3)-(r/R)`D. `(4 rho_(0)r)/(3epsilon_(0))(5/4)-(r/R)` |
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Answer» Correct Answer - C गोश प्रमेय के अनुशार `Exx4 pi r^(2) = (1)/(epsilon_(0)) overset(r )underset(0)int rho dV =(1)/(epsilon_(0)) overset(r )underset(0)int rho_(0) (5/4-(x)/(R )) 4 pi x^(2) dx` `therefore E=(1)/(4pi epsilon_(0)r^(2)) overset(r )underset(0)int 4pi rho_(0)(5/4 x^(2)-(1)/(R )x^(3))dx` `=(1)/(4piepsilon_(0)r^(2)) 4 pi rho_(0) [5/4 (r^(3))/(3)-(1)/(R ) (r^(4))/(4)]` `=(rho_(0)r)/(4 epsilon_(0))(5/3-(r )/(R))` अतः विकल्प C सही है |
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