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Equilibrium Constant and Standard Free Energy Change

A body of mas...

Question

A body of mass $1$ gram and carrying a charge $10^{- 8} C$ passes through two points P and Q. The electrostatic potential at Q is $0 V$ . The velocity of the body at Q is $0.2 m s^{- 1}$ and at P is $\sqrt{0.028} m s^{- 1}$ . The potential at P is:

150 V

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300 V

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600 V

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900 V

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Solution

The correct option is C $600 V$
We know $Δ$ v $\cdot Q = Δ k \cdot E$
$\Rightarrow (- v_{P} + v_{Q}) \times 10^{- 8} = \frac{1}{2} \times 1 \times 10^{- 3} ((\sqrt{0.028})^{2} - (0.2)^{2})$
$∴ - v_{P} \times 10^{- 8} = \frac{1}{2} \times 1 \times 10^{- 3} (0.028 - 0.04)$
$\Rightarrow V_{P} = \frac{10}{2} (0.012)$
$\Rightarrow V_{P} = 600 V$

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A body of mass 1 gram and carrying a charge 10−8C passes through two points P and Q. The electrostatic potential at Q is 0V. The velocity of the body at Q is 0.2ms−1 and at P is √0.028ms−1. The potential at P is:

The correct option is C 600VWe know Δ v⋅Q=Δk⋅E⇒(−vP+vQ)×10−8=12×1×10−3((√0.028)2−(0.2)2)∴−vP×10−8=12×1×10−3(0.028−0.04)⇒VP=102(0.012)⇒VP=600V

A body of mass $1$ gram and carrying a charge $10^{- 8} C$ passes through two points P and Q. The electrostatic potential at Q is $0 V$ . The velocity of the body at Q is $0.2 m s^{- 1}$ and at P is $\sqrt{0.028} m s^{- 1}$ . The potential at P is: