<p>A parallel-plate air capacitor of area A= 23.0 cm² and plate separ

Question

<p>A parallel-plate air capacitor of area A= 23.0 cm² and plate separation d= 3.20 mm is charged by a battery to a voltage 62.0 V. If a dielectric material with <span class="typeset"><span class="scale"><em>&amp;kappa</em></span></span> = 3.00 is inserted so that it fills the volume between the plates (with the capacitor still connected to the battery), how much additional charge will flow from the battery onto the positive plate?</p>

Explanation / Answer

Parallel plate cap C = e0er(A/d) in Farads e0 is 8.8542e-12 F/m er is dielectric constant (vacuum = 1) A and d are area of plate in m² and separation in m Initial capacitance is C = (8.85e-12)(23e-4)/(3.2e-3) = 63.60 pF Final cap = 3.0 x 63.6 pF = 190.82 pF Initial charge is Q = CV = 63.6 pF x 62v = 3943.2 pC

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