Energy stored in a Capacitor

1) A capacitance remains connected to a battery.If the plates of the capacitor are moved closer to each other what happens to a) Capacitance C b) Pottential  V c) Charge Q d) charge density σ e) Electric field E f) stored energy U                                                                                                                                                                                                                                                        (CBSE)

a) C= ε₀A / d

ie Cά 1 /d

as d, plate seperation decreases C increases

b) Since the battery is connected V remains the same

c) Q=CV and  V is a constant

ie Q ά C

so as C increases Q also increases ( it implies that charge flows from battery to capacitor)

d) σ=Q/ A

σ ά Q

as Q increases σ also increases

e) E= σ/ ε₀

E ά σ

as σ increases E also increases

or E=V/d as d decrease E increases

f) U=½ C V²

U ά C

as C increases U also increases

2) A capacitor is charged to a potential V using a battery This battery is removed and plates are moved close to each other then what happens to the parameters in problem 1)

Since the battery is removed after charging the charge Q  remains the same but V changes

Then use the above formulas and we can arrive at the following conclusions

C increases, V decreases, σ , Q and E constant , U decreases (U=½  Q² / C)