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 RU is a single capacitor. The charge on plate R is represented by Q_R. The capacitors are charged so that the potential (voltage) at A, V_A initially equals 15 volts. For each of the statements choose the proper response. If the plate separation for capacitor ST increases, the energy stored in RU will .... The voltage across capacitor ST is ... that across capacitor RU. Q_S is ... Q_R. The electric field between plates R and U is ... that between plates S and T. Q_S + Q_T is ... zero. The energy stored in capacitor ST is ... the energy stored in capacitor RU. If the plate separation for capacitor ST increases, the charge on Q_R 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)

If the plate seperation in Capacitor STincreases, the charge on QR will increase so energy stored in RU will increase

(2)

equal

(3)

Qs is greater than QR as Capacitance of ST is greater than RU as distance b/w plates is lesser in ST.

(4)

E = V/d , electric field is inversily proportional to distance between the plates, so, the electric field between plates R and U is less than that between plates S and T

(5)

Qs + QT is equal to Zero. becuase sum of positive and negative charges accumulated on opposite plates

(6)

E = 1/2 cV^2

capacitence of ST capacitor is more than RU because of less distance betwen the plates

so, the energy sotred in capacitor ST is greater than the energy stored in capacitor RU

(7)

If the plate seperation in Capacitor STincreases, the charge on QR will increase.

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