A parallel plate capacitor fitted with a material of dielectric constant 'k' is charged to a certain voltage. The dielectric material removed. Then ;-
a)capacitance reduces by a factor 'k'.
b) Electric field reduces by a factor 'k'.
c) Voltage across the capacitor increases by a factor 'k'.
d) charge stored in the capacitor increases by a factor 'k'.
options:
1) a and b 2) a and c 3) b and c 4) b and d
I know for sure capacitance reduces by a factor 'k' but not getting the combination right.
pl help

Question

A parallel plate capacitor fitted with a material of dielectric constant 'k' is charged to a certain voltage. The dielectric material removed. Then ;-
a)capacitance reduces by a factor 'k'.
b) Electric field  reduces by a factor 'k'.
c) Voltage across the capacitor increases by a factor 'k'.
d) charge stored in the capacitor increases by a factor 'k'.
options:
1) a and b 2) a and c  3) b and c  4) b and d
I know for sure capacitance reduces by a factor 'k' but not getting the combination right.
pl help

anonymous · Answer

2

assuming battery is also removed....charge remains same...hence voltage and field have to increase K times

anonymous · Answer

exactly right.. capacitance decreases.. and hence voltage increases!

anonymous · Answer

So option in this case would be serial 2 ( a&c). Thanks