The sketch below is a side view of two capacitors consisting of parallel plates

Question

The sketch below is a side view of two capacitors consisting of parallel plates in air. The capacitor plates are equal in area but the plate separation differs as shown. Individual capacitors are specified with two letters, for example SR is a single capacitor. The charge on plate S is represented by Qs. The capacitors are charged so that the potential (voltage) at A, VA, initially equals 29 volts. For each of the statements choose the proper response. The voltage across capacitor TU is . . . that across capacitor SR. QS + QR is . . . zero. QS is . . . QT. The electric field between plates S and R is . . . that between plates T and U. The energy stored in capacitor TU is . . . the energy stored in capacitor SR. If the plate separation for capacitor SR decreases, the energy stored in TU will If the plate separation for capacitor TU increases, the charge on QS will . . . . The electric field is proportional to the surface charge density on the plates and the plates of the two capacitors have the same area.

Explanation / Answer

1. Equal to 2. Qs + QR is equal to Zero. (sum of positive and negative charges accumulated on opposite plates) 3. Qs is less than QT as Capacitance of TU is greater than SR as distance b/w plates is lesser in TU. 4. Electric field b/w plates S and R is less than that of T and U 5. The energy stored in capacitor TU is more than that of SR 6. If the plate seperation in Capacitor SR decreases, the energy stored in TU decreases. 7. If the plate seperation in Capacitor TU increases, the charge on Qs will increase.

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

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