Name: Continuous Charge Distribution
Uploaded: 2019-07-19
Duration: 5 min 39 s
Description: This video will help us understand the necessity of continuous charge distribution and learn about the concept of linear charge density. Learn more.

Question

(i) Show that path of a charge particle projected normal to an uniform electrostatic field is a parabola.
(ii) A particle of mass

m

and charge

+ q

is thrown at a speed

u

against a uniform electric field

E

. How much distance will it travel before coming to rest?

Accepted Answer

(i) Force $+ q$ acting on a charged particle of charge $q$ in electric field $E$ is given by,

$F = (\frac{q}{m}) E$

If a charged particle moves in the direction of electric field, Then it is accelerated and will move in same direction of electric field.

But if a charged particle moves in a direction and not in parallel to electric field, it moves in a parabolic path.

Let $v$ be the velocity and $E$ be the electric field as shown in figure.

Let at $t = 0$ , y-component of velocity be $v_{y o}$ then vertical displacement y after a time duration t is given by,

$y = v_{y o} t + (\frac{1}{2}) a t^{2} . . . . . . . . . . . . . . . . . (1)$

Acceleration $a = \frac{q E}{m}$ and $t = \frac{x}{v_{x}}$ , where $x$ is horizontal displacement, $v_{x}$ component is unaffected by electric field

Hence, equation(1) is written as, $y = v_{y o} (\frac{x}{v_{x}}) + (\frac{1}{2}) a {(\frac{x}{v_{x}})}^{2} . . . . . . . . . . . . . . . . . . . . . . . (2)$

Vertical position $y$ as given by equation (1) is in parabolic curve of type $y = a x + b x^{2}$

(ii) Given,
$u =$ Velocity of projection

$E =$ Electric field intensity
$q =$ Charge on the particle

$m =$ Mass of particle
We know, Force experienced by a particle with charge ' $q$ ' in an electric field

$F = q E$

∴ The acceleration produced is given as:

$a = \frac{q E}{m} $

As the particle is projected against the electric field, hence deceleration $= \frac{q E}{m}$
So, let the distance covered be ' $s$ '
Then, $v^{2} = u^{2} + 2 a s$ [where $a$ =Acceleration, $v$ =Final velocity]
$0 = u^{2} - 2 \times \frac{q E}{m} \times s \Rightarrow s = \frac{u^{2} m}{2 q E} u n i t s$ .

(i) Show that path of a charge particle projected normal to an uniform electrostatic field is a parabola.
(ii) A particle of mass $m$ and charge $+ q$ is thrown at a speed $u$ against a uniform electric field $E$ . How much distance will it travel before coming to rest?

Important Questions on Electric Charges and Fields

Learn and Prepare for any exam you want !

(i) Show that path of a charge particle projected normal to an uniform electrostatic field is a parabola. (ii) A particle of mass m and charge +q is thrown at a speed u against a uniform electric field E. How much distance will it travel before coming to rest?

Important Questions on Electric Charges and Fields

Learn and Prepare for any exam you want !

(i) Show that path of a charge particle projected normal to an uniform electrostatic field is a parabola.
(ii) A particle of mass $m$ and charge $+ q$ is thrown at a speed $u$ against a uniform electric field $E$ . How much distance will it travel before coming to rest?