InterviewSolution
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InterviewSolution
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Plate A of parallel plate air filled capacitor is connected to a spring having force constant k and plate B is fixed. They are held on a frictionless table top as shown in Fig. If charge `+q` is placed on plate A and a charge `-q` on plate A and a charge `-q` on plate B, by how much does the spring expand ? |
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Answer» In a capacitor, plates carry equal and opposite charges. Therefore, there is a force of attraction between the plates. In a parallel plate capacitor, energy stored, `U = (1)/(2) (q^(2))/(C ) and C = (in_(0) A)/(x)` `:. U = (q^(2) x)/(2 in_(0) A)` As the electric field is conservation, therefore, As the electric field is conservative , therefore, `F = - (dU)/(dx) = - (d)/(dx) ((q^(2) x)/(2in_(0) A)) = - (q^(2))/(2 in_(0) A)` Negative sign imples that force is attractive. If k is force constant of the spring and it expands through a distance x, then `-F = kx` or `x = - (F)/(k) = (q^(2))/(2 in_(0) Ak)` |
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