A particle of mass 'm' and charge 'q' is fastened to one end 'A' of a massless string having equilibrium length , whose other end is fixed at point 'O'. The whole system is placed on a frictiionless horizontal plane and is initialy at rest. If uniform ele

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Q.
A particle of mass 'm' and charge 'q' is fastened to one end 'A' of a massless string having equilibrium length $l$ , whose other end is fixed at point 'O'. The whole system is placed on a frictionless horizontal plane and is initially at rest. If uniform electric field is switched on along the direction as shown in figure, then the speed of the particle when it crosses the x-axis is [2025]

1 $\sqrt{\frac{2 q E l}{m}}$

2 $\sqrt{\frac{q E l}{4 m}}$

3 $\sqrt{\frac{q E l}{m}}$

4 $\sqrt{\frac{q E l}{2 m}}$

Ans.
(3)

From work energy theorem,

Work done by electric force = change in kinetic energy or

$q E (l - l \cos 60 °) = \frac{1}{2} m v^{2}$

$\Rightarrow \frac{q E l}{2} = \frac{1}{2} m v^{2} or v = \sqrt{\frac{q E l}{m}}$

Solution

1 $\sqrt{\frac{2 q E l}{m}}$

2 $\sqrt{\frac{q E l}{4 m}}$

3 $\sqrt{\frac{q E l}{m}}$

4 $\sqrt{\frac{q E l}{2 m}}$

Ans.
(3)

From work energy theorem,

Work done by electric force = change in kinetic energy or

$q E (l - l \cos 60 °) = \frac{1}{2} m v^{2}$

$\Rightarrow \frac{q E l}{2} = \frac{1}{2} m v^{2} or v = \sqrt{\frac{q E l}{m}}$

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Solution

1 2qElm

2 qEl4m

3 qElm

4 qEl2m

Ans. (3) From work energy theorem, Work done by electric force = change in kinetic energy or qE(l–l cos 60°)=12mv2 ⇒ qEl2=12mv2 or v=qElm

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1 $\sqrt{\frac{2 q E l}{m}}$

2 $\sqrt{\frac{q E l}{4 m}}$

3 $\sqrt{\frac{q E l}{m}}$

4 $\sqrt{\frac{q E l}{2 m}}$

Ans.
(3)

From work energy theorem,

Work done by electric force = change in kinetic energy or

$q E (l - l \cos 60 °) = \frac{1}{2} m v^{2}$

$\Rightarrow \frac{q E l}{2} = \frac{1}{2} m v^{2} or v = \sqrt{\frac{q E l}{m}}$