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JEE PHYSICS MCQ: ELECTRIC POTENTIAL AND CAPACITANCE

1 Medium

1 of 1

Application +4 | -1

Capacitance of a parallel plate capacitor becomes

\frac{4}{3}

times its original value, if a dielectric slab of thickness

t = \frac{d}{2}

is inserted between the plates [

d

is the separation between the plates]. The dielectric constant of the slab is

View answer and explanation

Correct option: C

2

Option C (2) is correct because the dielectric constant is calculated using the formula for capacitance with a dielectric slab inserted.

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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