Back to ELECTRIC POTENTIAL AND CAPACITANCE

JEE PHYSICS MCQ: ELECTRIC POTENTIAL AND CAPACITANCE

1 Medium

1 of 1

Application +4 | -1

An ebonite plate

(\kappa=3), 6 \mathrm{~mm}

thick, is introduced between the parallel plates of a capacitor of plate area

2 \times 10^{-2} \mathrm{~m}^2

and plate separation

0.01 \mathrm{~m}

Find the capacitance.

View answer and explanation

Correct option: D

29.5 \mathrm{~pF}

Option D (29.5 pF) is correct because it is calculated using the formula for capacitance with a dielectric slab inserted between the plates.

More Questions from ELECTRIC POTENTIAL AND CAPACITANCE

Practice All Chapter MCQs →

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