Possible answers include \"equal to\" \"less than\" \"greater than\" \"increase\
ID: 1587015 • Letter: P
Question
Possible answers include "equal to" "less than" "greater than" "increase" "decrease" and "not change".
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 specificd with two letters, for example SR is a single capacitor. The charge on plate S is represented by Q_s. The capacitors are charged so that the potential (voltage) at A, V_A, initially equals 29 volts. For each of the statements choose the proper response. The energy stored in capacitor SR is... the energy stored in capacitor TU. Q_T + Q_U is... zero. Q_S is... Q_T. The voltage across capacitor SR is... that across capacitor TU. If the plate seperation for capacitor TU decreases, the energy stored in SR will.... The electric field between plates T and U is... that between plates S and R. If the seperation for capacitor SR decreases, the charge on Q_T will....Explanation / Answer
Solution:
The capacitors are connected in parallel.
So the voltage across the plates SR & TU is identical. V1=V2=V=29 volts.
The total charge = Q = qSR +qTU
The total energy in the circuit is conserved.
Since the separation of the plates is more in capacitor SR, its capacitance is LESS than that of capacitor TU
energy stored in a capacitor = 1/2 * C*V2
since VSR =VTU And CSR is LESS than CTU,
I) the energy stored in capacitor SR is LESS than that in TU
ii) QT + QU = 0 since the cahrge is distributed onthe two plates of the Same capacitor as +q & -q
iii) The charge on CSR < CTU ;QSR < QTU
iv) Since the capacitor plates are in parallel, VSR = VTU
v) Since the plate separation of TU decreases, its capacitance further Increases because E=1/2 CV2
As a result the energy stored on CSR will Decrease since thge total energy is conserved.
vi) Since electric field = E = kQ/d2 or E = V/d , ETU is greater than ESR
vii) If plate separation for SR decreases, the capacitance of SR will increase.
Since the system's charge is conserved, the charge on QTU will Decrease.