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Standard XII
Physics
Power in an AC Circuit
A parallel pl...
Question
A parallel plate capacitor of capacitance 100 microfarad is connected to a power supply of 200V. A di-electric slab of dielectric constant 5 is now inserted into gap. The work done by the supply during this process is?
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Q.
A parallel plate capacitor of capacitance
100
μ
F
is connected to a power supply of
200
V
. A dielectric slab of dielectric constant
5
is now inserted into the gap between the plates. The workdone by power supply will be:
Q.
A parallel-plate capacitor having plate area 400 cm
2
and separation between the plates 1⋅0 mm is connected to a power supply of 100 V. A dielectric slab of thickness 0⋅5 mm and dielectric constant 5⋅0 is inserted into the gap. (a) Find the increase in electrostatic energy. (b) If the power supply is now disconnected and the dielectric slab is taken out, find the further increase in energy. (c) Why does the energy increase in inserting the slab as well as in taking it out?
Q.
A parallel plate capacitor with a dielectric of dielectric constant
K
between the plates has a capacitance
C
and it is charged to
V
Volts
. The dielectric slab is slowly removed from the capacitor and then re-inserted between its plates. The net work done by the system in this process is:
Q.
A parallel-plate capacitor is connected across a battery. Now, keeping the battery connected, a dielectric slab is inserted between the plates. In this process,
Q.
A parallel-plate capacitor of capacitance 5 µF is connected to a battery of emf 6 V. The separation between the plates is 2 mm. (a) Find the charge on the positive plate. (b) Find the electric field between the plates. (c) A dielectric slab of thickness 1 mm and dielectric constant 5 is inserted into the gap to occupy the lower half of it. Find the capacitance of the new combination. (d) How much charge has flown through the battery after the slab is inserted?
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