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a ring of radius r is with uniformly distributed charge Q on it a charge q is now placed at the centre of the ring find the increment in tne tension in the ring?

r gabm , 8 Years ago
Grade 12th pass
anser 1 Answers
Vikas TU
Surface area of the  ring will be  = pi R^2
Hence the tension produced = Q * surface area
                                                     = Q *pi* R^2
Therfore, the increment in tne tension in the ring is in total Q *pi* R^2.
Last Activity: 8 Years ago
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