(numbers 11-13): An ideal parallel plate capacitor with a dielectric between the

Question

(numbers 11-13): An ideal parallel plate capacitor with a dielectric between the plates is charged and no charges may exit or enter the capacitor plates, which are fixed in position. When the dielectric material is removed:

11. The voltage between the plates: a) increases b) decreases c) remains the same d) more info is needed to determine

12. The electric field: a) increases b) decreases c) remains the same d) more info is needed to determine

13. The potential energy: a) increases b) decreases c) remains the same d) more info is needed to determine

please expain how you arrived to answer, thank you very much.

Explanation / Answer

11.

increases

Charge stored is given as

Q = CV

As the dielectric is removed , the capacitance decreases. Since the charge stored is constant , hence the Voltage across the plates increases to compensate for the decrease is capacitance.

12.

Increase

Electric field = Voltage/distance between plates

since the Voltage increase , and distance between plates does not change, hence the electric field also increase.

13.

PE = (0.5) Q2/C

Since C decreases , the PE increases   as PE is inversly related to C

Related Questions

Navigate

• Browse (All)
• Browse #
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.