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An electric capacitor consists of this round parallel plates. eachof radius R, separated by a distancel (l lt lt R) and uniformlycharged with surfacedensitiessigma and -sigma . Find the potentialof the electircfield and the magnitude of tis strengthvector at the axesof thecapacitor as functionsof a distanace x fromthe platesif x gt gt l. Investigateand obtainedexpressions at x gt gt R. |
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Answer» Solution :For `X gt 0` we can use the result as given above and write `varphi = (sigma l)/(2 epsilon_(0)) (1 - (|x|)/((R^(2) + x^(2))^(1//2)))` for the solution that solution that vanishes at `alpha` . There is a discontiunity in potentail for `|x| = 0`. The solution for NEGATIVE `x` is obtained by `sigma rarr -sigma`. THUS `varphi = (sigma//x)/(2 epsilon_(0) (R + x^(2))^(1//2))` + constant Hence ignorning the jump `E = - (DEL varphi)/(del x) = (sigma l R^(2))/(2epsilon_(0) (R^(2) + x^(2))^(3//2))` for large `|x| varphi ~~ +- (p)/(2pi epsilon_(0) x^(2))` and`E ~~ (p)/(2pi epsilon_(0)|x|^(3))` (where `p = pi R^(2) sigma l`) |
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