Question

Two parallel plates, each having area A 2625 em are connected to the terminals of a battery of voltage V6 6 V as shown. The plates are separated by a distance d-0.56 cm. 1) What is Q, the charge on the top plate? 2) What is U, the energy stored in the this capacitor? 3) The battery is now disconnected from the plates and the separation of the plates is doubled ( 1.12 cm). What is the energy stored in this new capacitor? 4) What is E, the magnitude of the electric field in the region between the plates? N/C

Answer 1

capacitanceC given A=2625cm" ;d=0.56 cm ->C-4.1504e -010F initial charge stored on the+ ve plate of the capacitor with voltageV=+6.0V Q-CV 2.4902e-009C CV 2)energy stored in capactitorU =7.4707e-009 J 3)if the seperation between the plates is double, the new capacitance C" 2.0752e-010F new energy stored in capacitor-U"- 2d 2 3.7353e-009 J 4) electric field between the plates E-1071.4N/C d dC voltage across plates