Question

..ooo AT&T 11:26 AM 1 8296.0 LonCAPA Homework 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 TU is a single capacitor. The charge on plate T is represented by QT. The capacitors are charged so that the potential (voltage) at A, VA, initially equals 17 volts. For each of the statements choose the proper response. V-0 less tharn The energy stored in capacitor TU is the energy stored in capacitor RS. QR Qs is zero. equal to decrease If the plate separation for capacitor greater thanThe electric field between plates T les thanQR Qr- RS decreases, the energy stored in TU will . and U is that between plates R and S. decrease If the plate separation for capacitor S decreases, the charge on Qr wil equal to The voltage across capacitor RS is

Answer 1

The capacitance of a parallel plate capacitor is given by: $C=\frac{\epsilon _{0}*A}{d}$ where d is the seperation between the plates.Also, $Q=C*V$ where Q is the charge stored in the capacitor.

The energy stored in the capacitor is given by: $E=\frac{1}{2}*C*V^2\, \, or\, \, \frac{Q^2}{2*C}$ .

1)

The potential difference across both the capacitors is equal by the Kirchoff's voltage law. As the plate seperation in capacitor TU is more, its capacitance is less by the above equation.Hence by the expression for energy, energy stored in TU is less than RS.

2)

The charge on the plates of the capacitor is opposite and equal in magnitude.Thus, Q_R+Q_S is always zero.

3)

The plate T and R are connected to each other only and hence the sum of the charges on these plates must remain constant.Thus, the sum of charges on the 2 capacitors must remain constant.Moreover, the potential difference across them is also constant.

When plate seperation of RS decreases, its capacitance increases.Thus, the potential across the capacitor RS decreases.To maintain constant potential , charge flows from plate T to R and effectively increasing the charge on capacitor RS and reducing it on capacitor TU. Finally, using equation E=Q²/2*C , we can conclude that the energy stored in the capacitor TU decreases.

4)

The Electric field is related to the potential by:   $E=-\frac{\mathrm{d} V}{\mathrm{d} r}$ .For the parallel plate capacitors the field in the gap can be assumed to be uniform and in the direction normal to the plate. So, $E=-\frac{\Delta V}{\Delta r}=-\frac{\Delta V}{d}$

as the potential difference across both the capacitors is equal, the Electric field between the plates T and U is less than that of R and S , as the gap is more in TU.

5)

The capacitance of TU is less , so the charge stored in TU is also less than that in RS ,by the equation Q=C*V, as the potential across the capacitors is equal.

6)

If the plate seperation for capacitor RS decreases, the charge on capacitor TU decreases as explained before.Hence, Q_T decreases.