a parallel plate capacitor filled with air is connected to abattery and becomes

Question

a parallel plate capacitor filled with air is connected to abattery and becomes fully charged. the capacitor is thendisconnected and the seperation between the plates isdoubled. assuming that no charge leaks off, determine how thefollowing five quantities are affected, specify wether it increasesor decreases and by what factor. a) capacitance b) voltage across the capacitors plates c) electric field inside the capacitor d)change on the plates of the capacitor e)energy stored in the capacitor a parallel plate capacitor filled with air is connected to abattery and becomes fully charged. the capacitor is thendisconnected and the seperation between the plates isdoubled. assuming that no charge leaks off, determine how thefollowing five quantities are affected, specify wether it increasesor decreases and by what factor. a) capacitance b) voltage across the capacitors plates c) electric field inside the capacitor d)change on the plates of the capacitor e)energy stored in the capacitor

Explanation / Answer

a) C = 0 A / d     so if d is doubled the capacitance is halved An important relation is Gauss Law Q =0 = 0 E A b) The potential V = E d / Q (work / charge) = (Q/ 0 A) d / Q = d / (0 A)      so if d is doubled V isdoubled c) From above and by Gauss Law E remainsunchanged d) The charge remains constant as there is no currentflow e) E = 1/2 C V2    ............    = 1/2 * (1 / 2)* 22   = twice the originalenergy     (energy is required to move the platesapart)

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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