In a uniform electric field if a charge is fired in a direction different from the line of electric field then the trajectory of the charge will be a

Question

In a uniform electric field if a charge is fired in a direction different from the line of electric field then the trajectory of the charge will be a (A) Straight line (B) Circle (C) Parabola (D) Ellipse

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/d5c9ad01149179e6f041d3a4de715ed9-.png" /> F=q E=m ax ax=[(q E/m)] Then, x=0+(1/2) ax t2 ...(i) But y=ux t then t=(y/u x) So, x=(1/2) (q E/m) ⋅ (y2/u2) so x propto y2 for parabola

Q. In a uniform electric field if a charge is fired in a direction different from the line of electric field then the trajectory of the charge will be a

Straight line

Circle

Parabola

Ellipse

$F = qE = m a_{x}$
$a_{x} = [\frac{qE}{m}]$
Then, $x = 0 + \frac{1}{2} a_{x} t^{2} ... (i)$
But $y = u_{x} t$ then $t = \frac{y}{ux}$
So, $x = \frac{1}{2} \frac{qE}{m} \cdot \frac{y ^{2}}{u ^{2}}$
so $x \propto y^{2}$ for parabola

Q. In a uniform electric field if a charge is fired in a direction different from the line of electric field then the trajectory of the charge will be a

Straight line

Circle

Parabola

Ellipse

F=qE=max​ ax​=[mqE​] Then, x=0+21​ax​t2...(i) But y=ux​t then t=uxy​ So, x=21​mqE​⋅u2y2​ so x∝y2 for parabola

$F = qE = m a_{x}$
$a_{x} = [\frac{qE}{m}]$
Then, $x = 0 + \frac{1}{2} a_{x} t^{2} ... (i)$
But $y = u_{x} t$ then $t = \frac{y}{ux}$
So, $x = \frac{1}{2} \frac{qE}{m} \cdot \frac{y ^{2}}{u ^{2}}$
so $x \propto y^{2}$ for parabola