QJEEM22Ph8f8a: JEE Main 2022 Physics - Class 12 JEEM22Ph8f8a Solution: A condenser of 2 μ F capacitance is charged steadily from 0 to…

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QJEEM22Ph8f8a: JEE Main 2022 Physics - Class 12 JEEM22Ph8f8a Solution: A condenser of 2 μ F capacitance is charged steadily from 0 to…

A condenser of $2 μ F$ capacitance is charged steadily from 0 to $5 C$ . Which of the following graph represents correctly the variation of potential difference $(V)$ across it's plates with respect to the charge $(Q)$ on the condenser?

Multiple Choice Options

A https://ik.imagekit.io/novatra/photos/668a639894f424697296856234637bf5b4759297f635b96efdf41c05d625bbff.webp
B https://ik.imagekit.io/novatra/photos/91663a7dad6c8e1694031aedec77ede2ad022e26d24c855c00a51de9e68c4e94.webp
C https://ik.imagekit.io/novatra/photos/f73dbffbc127193964ac9846ad9c2bfc120f5b9d7f157646c45ca13208114d85.webp
D https://ik.imagekit.io/novatra/photos/58db5a209595d4e07eca82cb80fee5e07a408a7144fa60f2949579a091bde9ef.webp

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Accepted Answer

Q = CV

As capacitance is constant so, Q $\propto$ V

So graph between V and Q will be a straight line.

Initially voltage, $V_{i} = \frac{Q_{i}}{C} = \frac{0}{2 \times 10^{- 6}} = 0 V$

and $V_{f} = \frac{Q_{f}}{C} = \frac{5}{2 \times 10^{- 6}} = 2.5 \times 10^{6} V$

So correct graph will be A.

QJEEM22Ph8f8a: JEE Main 2022 Physics - Class 12 JEEM22Ph8f8a Solution: A condenser of 2 μ F capacitance is charged steadily from 0 to…

Multiple Choice Options

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