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Electric Field Due to a Dipole Along the Equatorial Line

A particle of...

Question

A particle of mass $m$ and charge $q$ is projected from the origin with a velocity $v_{0}^j$ in a region where an electric field is given by $\to E = E_{0}^i$ , at time $t = 0$ . Then choose the correct statement(s).(Given $m v_{0}^{2} = 2 q E_{0} x_{0}$ )

Radius of curvature of the particle when its x coordinate becomes

x_{0}

2 x_{0}

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Radius of curvature of the particle when its x coordinate becomes

x_{0}

4 \sqrt{2} x_{0}

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Speed of the particle when its x-coordinate becomes

x_{0}

\sqrt{2} v_{0}

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Speed of the particle when its x-coordinate becomes

x_{0}

2 v_{0}

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Solution

The correct options are
B Radius of curvature of the particle when its x coordinate becomes $x_{0}$ is $4 \sqrt{2} x_{0}$ .
C Speed of the particle when its x-coordinate becomes $x_{0}$ is $\sqrt{2} v_{0}$ .
$u = v_{0}^j$
$E = E_{0}^i$
$\to a = \frac{q E_{0}}{m}^i$
Radius of curvature $R = \frac{v^{2}}{a_{⊥}}$
velocity along x-axis, when $x = x_{0}$
$v_{x} = \sqrt{2 a x_{0}} = \sqrt{\frac{2 q E_{0} x_{0}}{m}} = v_{0}$

As there is no acceleration along y axis,
$v_{y} = v_{0}$

Net speed $= \sqrt{v_{x}^{2} + v_{y}^{2}} = v_{0} \sqrt{2}$
$R = \frac{(v_{0} \sqrt{2})^{2}}{a cos 45^{\circ}} = \frac{m (v_{0} \sqrt{2})^{2} \sqrt{2}}{q E_{0}} = \frac{m v_{0}^{2} (2 \sqrt{2})}{q E_{0}}$
$R = \frac{2 q E_{0} x_{0} (2 \sqrt{2})}{q E_{0}} = 4 \sqrt{2} x_{0}$

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