Two parallel plates, each having area A = 3391cm2 are connected to the terminals

Question

Two parallel plates, each having area A = 3391cm2 are connected to the terminals of a battery of voltage Vb = 6 V as shown. The plates are separated by a distance d = 0.41cm. You may assume (contrary to the drawing) that the separation between the plates is small compared to a linear dimension of the plate. A dielectric having dielectric constant ? = 3.2 is now inserted in between the plates of the capacitor as shown. The dielectric has area A = 3391 cm2 and thickness equal to half of the separation (= 0.205 cm) . What is the charge on the top plate of this capacitor? What is U, the energy stored in this capacitor? The battery is now disconnected from the capacitor and then the dielectric is withdrawn. What is V, the voltage across the capacitor?

Explanation / Answer

fiest calculate C= EoA/ d-t (1-1/k) ..here d= distance between plates ... t = thickness of slab k = relative permittivity of the dielectric material between the plates...... Then Q = C* V.... Energy stored = (1/2)CV^2.... after slab is removed C' = EoA/d. Charge on the plate remain same when battery is removed. So V = Q/C'......... I hope u can do mathematical calculation..... Please rate

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