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JEE PHYSICS MCQ: ELECTRIC POTENTIAL AND CAPACITANCE

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Two equal charges ‘q’ of opposite sign are separated by small distance ‘2a’. The electric intensity ‘E’ at a point on the perpendicular bisector of the line joining the two charges at a very distance ‘r’ from the line is

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Correct option: B

\frac{1}{{4\pi {\varepsilon _0}}}\,\frac{{2qa}}{{{r^3}}}

Option B is correct because the electric field due to a dipole at a point on the perpendicular bisector is inversely proportional to the cube of the distance.

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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