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JEE PHYSICS MCQ: ELECTRIC POTENTIAL AND CAPACITANCE

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Application +4 | -1

A parallel plate capacitor of capacitance 5 micro farad is charged to

120 \mathrm{~V}

and then connected to another uncharged capacitor. If the potential falls to

40 \mathrm{~V},

the capacitance of the second capacitor is

View answer and explanation

Correct option: C

15 micro farad

Option C (15 micro farad) is correct because the charge is conserved when the capacitors are connected, allowing us to calculate the capacitance of the second capacitor.

More Questions from ELECTRIC POTENTIAL AND CAPACITANCE

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The potential at a point distance

x

(measured in

\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