A particle with charge e and mass m, moving along the X-axis with a uniform speed u, enters a region where a uniform electric field E is acting along the Y-axis. The particle starts to move in a parabola. Its focal length (neglecting any effect of gravity) is

Question

A particle with charge e and mass m, moving along the X-axis with a uniform speed u, enters a region where a uniform electric field E is acting along the Y-axis. The particle starts to move in a parabola. Its focal length (neglecting any effect of gravity) is (A) (2mu2/eE) (B) (eE/2mu2) (C) (mu/2eE) (D) (mu2/2eE)

Tardigrade Admin · Accepted Answer

From parabola <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/fb908276d12e5b8863bbe97171a926ae-.png" /> y=(1/2) ×[(E e/m)] × (X2/u2) =(E ⋅ e/2 m u2) ⋅ X2 As X=4 a y (for parabola) ∴ X2 =(2 m u2/E ⋅ e) ⋅ y a =(2 m u2/4 E ⋅ e)=(m u2/2 E ⋅ e)

Q. A particle with charge e and mass $m$ , moving along the $X$ -axis with a uniform speed $u$ , enters a region where a uniform electric field $E$ is acting along the $Y$ -axis. The particle starts to move in a parabola. Its focal length (neglecting any effect of gravity) is

$\frac{2 m u ^{2}}{e E}$

$\frac{e E}{2 m u ^{2}}$

$\frac{m u}{2 e E}$

$\frac{m u ^{2}}{2 e E}$

From parabola

$y = \frac{1}{2} \times [\frac{E e}{m}] \times \frac{X ^{2}}{u ^{2}}$
$= \frac{E \cdot e}{2 m u ^{2}} \cdot X^{2}$
As $X = 4 a y$ (for parabola)
$∴ X^{2} = \frac{2 m u ^{2}}{E \cdot e} \cdot y$
$a = \frac{2 m u ^{2}}{4 E \cdot e} = \frac{m u ^{2}}{2 E \cdot e}$

Q. A particle with charge e and mass m, moving along the X-axis with a uniform speed u, enters a region where a uniform electric field E is acting along the Y-axis. The particle starts to move in a parabola. Its focal length (neglecting any effect of gravity) is

eE2mu2​

2mu2eE​

2eEmu​

2eEmu2​