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JEE PHYSICS MCQ: ELECTRIC POTENTIAL AND CAPACITANCE

1 Easy

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Application +4 | -1

A capacitor of capacitance

C_1

is charged to a potential

V

and then connected in parallel to an uncharged capacitor of capacitance

C_2

. The final potential difference across each capacitor will be

View answer and explanation

Correct option: A

\frac{C_1 V}{C_1+C_2}

Option A (

\frac{{C_1 V}}{{C_1+C_2}}

) is correct because the final potential difference is determined by the charge conservation and the total capacitance of the parallel combination.

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The potential at a point distance

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\mu \mathrm{m}

) due to some charges situated on the

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V(x)=\frac{20}{x^2-5} \mathrm{V}

. The electric field at

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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.
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E

will have minimum potential energy, if the positive direction of dipole moment makes the following angle with

E