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JEE PHYSICS MCQ: ELECTRIC POTENTIAL AND CAPACITANCE

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Application +4 | -1

A capacitor of unknown capacitance

\mathrm{C}

is connected across a battery of

\text{V}

volt. The charge stored in it becomes

\text{Q}

coulomb. When potential across the capacitor is reduced by

\text{V}^{\prime}

volt, the charge stored in it becomes

\text{Q}^{\prime}

coulomb. The capacitance

\text{C}

is

View answer and explanation

Correct option: A

\frac{\mathrm{Q}-\mathrm{Q}^{\prime}}{\mathrm{V}^{\prime}}

Option A (

\frac{\mathrm{Q}-\mathrm{Q}^{\prime}}{\mathrm{V}^{\prime}}

) is correct because the capacitance is calculated using the change in charge and the change in voltage across the capacitor.

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\mu \mathrm{m}

) due to some charges situated on the

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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

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. 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

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. 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

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