A parallel plate capacitor with an air gap is connected to a 6V battery. The charged...

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A parallel plate capacitor with an air gap is connected to a 6V battery. The charged capacitor stores energy of 72nJ. Without disconnecting the capacitor from the battery, a dielectric material is inserted into the air gap and the energy stored in the capacitor is increased to 317nJ. A conducting plate in the capacitor has an area of 50 cm². (a) What is the charge on the positive plate of the capacitor after the dielectric is inserted? (b) What is the magnitude of the electric field between the plates before the dielectric is inserted? (c) What is the dielectric constant of the material? A parallel plate capacitor with an air gap is connected to a 6V battery. The charged capacitor stores energy of 72nJ. Without disconnecting the capacitor from the battery, a dielectric material is inserted into the air gap and the energy stored in the capacitor is increased to 317nJ. A conducting plate in the capacitor has an area of 50 cm². (a) What is the charge on the positive plate of the capacitor after the dielectric is inserted? (b) What is the magnitude of the electric field between the plates before the dielectric is inserted? (c) What is the dielectric constant of the material?

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a The energy stored in a capacitor can be calculated using the formula E 12QV Where E energy stored ... View the full answer