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Byju's Answer

Standard XII

Physics

Potential Energy of a System of Multiple Point Charges

In the regin ...

Question

In the regin electric field exists along x-axis, the electric potential along x-axis varies as shown in the graph. A charge particle of charge $q$ and mass $m$ is released at $x = 0.5 m$ . The maximum velocity achieved by the charge particle is:

\sqrt{\frac{q V_{0}}{m}}

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\sqrt{\frac{2 q V_{0}}{m}}

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\sqrt{\frac{q V_{0}}{2 m}}

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2 \sqrt{\frac{V_{0}}{2 m}}

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Solution

The correct option is B $\sqrt{\frac{2 q V_{0}}{m}}$
Given,
Charge $= q$
Mass $= m$
Released $x = 0.5 m$
Maximum velocity for equilbrium condition
Then $E = \frac{1}{2} m v^{2} \frac{1}{2} m v^{2} = q V_{0} V = \sqrt{\frac{2 q V_{0}}{m}}$

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In the regin electric field exists along x-axis, the electric potential along x-axis varies as shown in the graph. A charge particle of charge q and mass m is released at x=0.5m. The maximum velocity achieved by the charge particle is:

The correct option is B √2qV0mGiven,Charge=qMass=mReleasedx=0.5mMaximum velocity for equilbrium conditionThen E=12mv212mv2=qV0V=√2qV0m

In the regin electric field exists along x-axis, the electric potential along x-axis varies as shown in the graph. A charge particle of charge $q$ and mass $m$ is released at $x = 0.5 m$ . The maximum velocity achieved by the charge particle is:

The correct option is B $\sqrt{\frac{2 q V_{0}}{m}}$
Given,
Charge $= q$
Mass $= m$
Released $x = 0.5 m$
Maximum velocity for equilbrium condition
Then $E = \frac{1}{2} m v^{2} \frac{1}{2} m v^{2} = q V_{0} V = \sqrt{\frac{2 q V_{0}}{m}}$