A. 25 V
B. 12.5 V
C. 10 V
D. 22.5 V
A 2μF condenser is charged to 500 V and then the plates are joined through a resistance in joule is
A dielectric of thickness 5cm and dielectric constant 10 is introduced between the plates of a parallel plate capacitor having plate area 500 sq.cm and separation between the plates 10cm.The capacitance of the capacitor with dielectric slab is(ε0=8.8x10-12/N-m2)
A 4μF capacitor is charged by a 200V battery.It is then disconnected from the supply and is connected to another uncharged 2μF capacitor.During this process,Loss of energy(in J)is
a spark is produced between two insulated surfaces maintained at a potential difference of 5 x 106 volt. The energy output is 10-5 J. The charge transferred during the spark is
A capacitor of 8 μF is charged to a potential of 1000V. The energy stored in the capacitor is
A 100 PFcapacitor is charged to a potential difference of 24V.It is connected to an uncharged capacitor 20PF.what will be the new potential difference across 100PF capacitor
Charge 20,30,-40 and 50 μC are at the corners of a square of 10cm. The field at the point of intersection of the diagonals is
Three condensers 1μF,2μF and 3μF are connected in series to a p.d of 330volt.The PD across the plates of 3μF is
Two parallel plate air capacitors have the same separation. The plates of the first are squares of side 10cm.The plates of the second are squares of side 20cm.The ratio of the capacities respectively is
The capacity of a parallel plate condenser consisting of two plates each 10 cm square and are separated by a distance of 2mm is (Take air as the medium between the plate):