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JEE PHYSICS MCQ: ELECTRIC POTENTIAL AND CAPACITANCE

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Analyzing +4 | -1

A parallel plate capacitor is charged by
connecting a 2 V battery across it. It is then
disconnected from the battery and a glass slab
is introduced between plates. Which of the
following pairs of quantities decrease?

View answer and explanation

Correct option: D

Potential difference and energy stored

Option D (Potential difference and energy stored) is correct because when the battery is disconnected, the charge remains constant, and introducing a glass slab increases capacitance, thus decreasing potential difference and energy stored.

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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