Two experiments are carried out with a parallel plate capacitor (a) The capacito

Question

Two experiments are carried out with a parallel plate capacitor (a) The capacitor is charged to a potential difference of 100 V, disconnected from the source, and then a slab of dielectric is inserted between the plates. (b) The capacitor is connected to a battery which maintains a potential difference of 100 V between the capacitor plates and then a slab of dielectric is inserted between the plates. In which experiments is the energy stored by the capacitor greatest? experiment (a) experiment (b) both the same need more information

Explanation / Answer

In first case the charge on capacitor will remain same. Capacitance will increase. Hence by E = 1/2*Q2/C Energy stored will decrease.

In second case Voltage remains same but capacitance increase hence by E = 1/2*CV2 Energy will increase.

Hence option(B) is correct as energy stored in second capacitor is greatest.

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

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