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Unformatted text preview: Question Orange 1.1) First of all, we use the Gausss law for a single plate to obtain the electric field, = E . (0.2) The density of surface charge for a plate with charge, Q and area, A is A Q = . (0.2) Note that the electric field is resulted by two equivalent parallel plates. Hence the contribution of each plate to the electric field is E 2 1 . Force is defined by the electric filed times the charge, then we have Force = Q E 2 1 = A Q 2 2 (0.2)+ (0.2) ( The coefficient + the final result) 1.2) The Hooks law for a spring is x k F m = . (0.2) In 1.2 we derived the electric force between two plates is A Q F e 2 2 = . The system is stable. The equilibrium condition yields e m F F = , (0.2) k A Q x 2 2 = (0.2) 1.3) The electric field is constant thus the potential difference, V is given by ) ( x d E V = (0.2) (Other reasonable approaches are acceptable. For example one may use the definition of capacity to obtain V .) By substituting the electric field obtained from previous section to the above equation, we get,  = d k A Q A d Q V 2 2 1 (0.2) 1.4) C is defined by the ratio of charge to potential difference, then V Q C = . (0.1) Using the answer to 1.3, we get Using the answer to 1....
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 Spring '11
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