JEE Main 2020

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Animated Solution for Physics - Electrostatics: A particle of mass $m$ and charge $q$ is released from rest in a uniform electric field. If there is no other force on the particle, the dependence of its speed $v$ on the distance $x$ travelled by it is correctly given by (graphs are schematic and not drawn to scale)

Q: Why can't we use $v = u + at$ to find the graph?

While $v = u + at$ is perfectly valid, it gives the relationship between velocity and *time*. The question specifically asks for the dependence of speed $v$ on the *distance* $x$. Therefore, $v^2 - u^2 = 2ax$ is the direct and appropriate choice.

Q: Why does the graph curve downwards instead of upwards?

The equation is $v^2 \propto x$, which means $v \propto \sqrt{x}$. As $x$ increases, $v$ increases, but the *rate* of increase slows down. This creates a parabola that opens sideways along the x-axis, appearing to curve 'downwards' relative to a straight line.

Select Answer:

Visualized Solution

$I ni t ia l S e t u p$

Particle of mass $m$ and charge $q$ is released from rest ( $u = 0$ ) in a uniform electric field $E$ .

$E l ec t r i c F or ce$

$F = q E$

$A cce l er a t i o n$

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

$E q u a t i o n o f M o t i o n$

$v^{2} - u^{2} = 2 a x$

$S u b s t i t u t in g V a l u es$

$v^{2} - 0^{2} = 2 (\frac{q E}{m}) x$

$V e l oc i t y - D i s t an ce R e l a t i o n$

$v^{2} = (\frac{2 q E}{m}) x$

$G r a p h I d e n t i f i c a t i o n$

$v^{2} \propto x ⟹ Parabola opening along +ve x-axis$

$C o n c l u s i o n$

$The correct graph is (c).$

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The Sigma Insight: Electric Field

Solution Diagram

The Physical Setup

Imagine you are observing a tiny particle with mass

m

and charge

q

, initially completely at rest. Suddenly, we switch on a uniform electric field, denoted by

E

Because the particle carries a charge, this electric field immediately exerts an electrostatic force on it. According to Coulomb's law for an electric field, this force is given by

F = q E

Since the electric field is uniform, this force is perfectly constant in both magnitude and direction. It acts as a steady, invisible hand pushing the particle forward.

The Master Equation

Now, we must connect this force to the particle's motion. Newton's second law tells us that a net force produces an acceleration,

a = \frac{F}{m}

Substituting our electric force, we find the acceleration to be

a = \frac{q E}{m}

This is a crucial realization: because

q

E

, and

m

are all constants, the acceleration

a

is also strictly constant. This allows us to unlock the powerful equations of uniformly accelerated motion!

Final Calculation

We want to find how the particle's speed

v

depends on the distance

x

it has traveled. The perfect tool for this is the third equation of kinematics:

v^{2} - u^{2} = 2 a x

Since the particle was released from rest, its initial velocity

u

is exactly zero. Substituting

u = 0

and our constant acceleration

a = \frac{q E}{m}

, the equation transforms into:

v^{2} = 2 (\frac{q E}{m}) x

Look closely at the structure of this final result. The entire term in the parentheses,

\frac{2 q E}{m}

, is just a constant number. Let's call it

K

. Our equation simplifies to

v^{2} = K x

Mathematically, this is the classic equation of a parabola (

y^{2} = 4 a x

) that opens along the positive x-axis. As the distance

x

increases, the speed

v

also increases, but at a gradually decreasing rate. This perfectly matches the curve shown in option (c), making it our correct answer!

Common Traps & FAQs

Why can't we use

v = u + a t

to find the graph?

While

v = u + a t

is perfectly valid, it gives the relationship between velocity and *time*. The question specifically asks for the dependence of speed

v

on the *distance*

x

. Therefore,

v^{2} - u^{2} = 2 a x

is the direct and appropriate choice.

How do we know the acceleration is constant?

The problem states the electric field is 'uniform'. A uniform field exerts a constant force (

F = q E

) on a constant charge. By Newton's second law (

a = F / m

), a constant force on a constant mass results in a constant acceleration.

Why does the graph curve downwards instead of upwards?

The equation is

v^{2} \propto x

, which means

v \propto x

. As

x

increases,

v

increases, but the *rate* of increase slows down. This creates a parabola that opens sideways along the x-axis, appearing to curve 'downwards' relative to a straight line.

What if the particle had a negative charge?

If the charge were negative, the force and acceleration would be in the opposite direction of the electric field. However, if we define