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Standard XII

Physics

How to Visualize Electric Fields

If a mass of ...

Question

If a mass of $20 g$ having charge $3.0 m C$ moving with velocity $20 m s^{- 1}$ enters a region of electric field of $80 N C^{- 1}$ in the same direction as the velocity of mass, then the velocity of mass after $3 s$ in the region will be

40 m s^{- 1}

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44 m s^{- 1}

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56 m s^{- 1}

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80 m s^{- 1}

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Solution

The correct option is D $56 m s^{- 1}$
As the mass is moving in the electric field, then,
$m a = q E$
$a = \frac{q E}{m} = \frac{3 \times 10^{- 3} \times 80}{20 \times 10^{- 3}} = 12 m s^{- 2}$
By using, $v = u + a t$
$v = 20 + 12 \times 3 = 56 m s^{- 1}$ .

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If a mass of 20 g having charge 3.0 mC moving with velocity 20 ms−1 enters a region of electric field of 80 NC−1 in the same direction as the velocity of mass, then the velocity of mass after 3s in the region will be

The correct option is D 56 ms−1As the mass is moving in the electric field, then,ma=qEa=qEm=3×10−3×8020×10−3=12ms−2By using, v=u+atv=20+12×3=56 ms−1.

If a mass of $20 g$ having charge $3.0 m C$ moving with velocity $20 m s^{- 1}$ enters a region of electric field of $80 N C^{- 1}$ in the same direction as the velocity of mass, then the velocity of mass after $3 s$ in the region will be

The correct option is D $56 m s^{- 1}$
As the mass is moving in the electric field, then,
$m a = q E$
$a = \frac{q E}{m} = \frac{3 \times 10^{- 3} \times 80}{20 \times 10^{- 3}} = 12 m s^{- 2}$
By using, $v = u + a t$
$v = 20 + 12 \times 3 = 56 m s^{- 1}$ .