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Find the force of attraction between the plates of a parallel plate capacitor. |
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Answer» Solution :Let d be the distance between the plates. Then the capacitor is C = `(epsilon_(0)A)/(d)` Energy STORED in a capacitor, U = `(q^(2))/(2C) = (q^(2).d^(2))/(2 epsilon_(0)A) ` Energy MAGNITUDE of the force is, `|F| = (dU)/(dx) = (d)/(dx) [ (q^(2) x)/(2 epsilon_(0) A) ] "" [ x = d ]` = `(q^(2))/(2 epsilon_(0)A) [ (d)/(dx) (x) ] ` F = `(q^(2))/(2 epsilon_(0)A)` |
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