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A non-conducting semi circular disc (as shown in figure) has a uniform surface charge density `sigma`. The ratio of electric field to electric potential at the centre of the disc will be A. `(1)/(pi)(lnb//a)/((b-a))`B. `(2)/(pi)`C. `(1)/(pi)(ln(b//a)^(2))/((b-a))`D. `(pi(b-a))/(2ln(b//a))` |
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Answer» Correct Answer - C `E=intdE= underset(x=a)overset(b)(int)(2K sigma dx)/(x)=2 K sigma l n(b)/(a)` `V = int dv = underset(x=1)overset(b)(int) (K pi x sigma dx)/(x) = K sigma pi (b-a)` `rArr (E)/(V)=(2 k sigmal n(b//a))/(k sigma pi(b-a))=(ln (b//a)^(2))/(pi(b-a))`. |
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