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JEE PHYSICS MCQ: ELECTRIC POTENTIAL AND CAPACITANCE

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Application +4 | -1

A parallel-plate capacitor of capacitance

100 \mu \mathrm{F}

is connected to a power supply of

200 \mathrm{~V}

. A dielectric slab of dielectric constant 5 is now inserted into the gap between the plates. (a) Find the extra charge flown through the power supply and the work done by the supply. (b) Find the change in the electrostatic energy of the electric field in the capacitor.

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Correct option: A

16 J; 8 J

Option A (16 J; 8 J) is correct because the extra charge flown through the power supply and the work done by the supply are 16 J, and the change in the electrostatic energy of the electric field in the capacitor is 8 J.

More Questions from ELECTRIC POTENTIAL AND CAPACITANCE

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The potential at a point distance

x

\mu \mathrm{m}

) due to some charges situated on the

x

axis is given by

V(x)=\frac{20}{x^2-5} \mathrm{V}

. The electric field at

x=5 \mu \mathrm{m}

is given by: A parallel plate capacitor is connected to a 5V battery and charged. The battery is then disconnected and a glass slab is introduced between the plates. Then, the quantities that decrease are A conducting sphere of radius

R

is given a charge

Q

. The electric potential and the electric field at the centre of the sphere respectively are Two capacitors when connected in series have a capacitance of

3\mu F,\,

and when connected in parallel have a capacitance of

16\mu F

. Their individual capacities are Statement-1: If temperature is increased, the dielectric constant of a polar dielectric decreases whereas that of a non-polar dielectric does not change significantly.
Statement-2: The magniude of dipole moment of individual polar molecule decreases significantly with increase in temperature. An electric dipole when placed in a uniform electric field

E

will have minimum potential energy, if the positive direction of dipole moment makes the following angle with

E