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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
 one year ago
 one year ago
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
 one year ago
 one year ago

This Question is Closed

telltoamitBest ResponseYou've already chosen the best response.2
2 assuming battery is also removed....charge remains same...hence voltage and field have to increase K times
 one year ago

MashyBest ResponseYou've already chosen the best response.0
exactly right.. capacitance decreases.. and hence voltage increases!
 one year ago

haridas_mandalBest ResponseYou've already chosen the best response.0
So option in this case would be serial 2 ( a&c). Thanks
 one year ago
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