Question

There is a uniform electric field of strength ${{10}^{3}}V{{m}^{-1}}$ along the y-axis. A body of mass 1g and charge ${{10}^{-6}}C$ is projected into the field from origin along the positive x-axis with a velocity $10m{{s}^{-1}}$. Its speed in $m{{s}^{-1}}$ after 10s is (neglect gravitation) :
A. $10$
B. $5\sqrt{2}$
C. $10\sqrt{2}$
D. $20$

Answer 1

With the knowledge of electric field, force and Newton’s laws find the acceleration of the body. Then resolve the velocity of the body into x and y directions. Use the required kinematic equation and find the velocities of the body in x and y direction after 10 seconds.

Formula used:
$F=qE$
$F=ma$
$v=u+at$

Complete step by step answer:
It is given that there is a uniform electric field along the y-axis. The magnitude of the electric field is equal to ${{10}^{3}}V{{m}^{-1}}$. Then a body of some mass and some charge is projected into the field. The point of projection is at the origin in the direction along the positive x-axis. It is said that the body is projected with a velocity of $10m{{s}^{-1}}$.

Since it is said to neglect gravity, the only force that will affect the motion of the body is the electric force due to the electric field. The magnitude of electric force exerted on a charge particle q placed in an electric field E is equal to $F=qE$ …. (i).

The direction of the electric force is the same as the direction of the electric field.
In this case, the electric field is in the direction of the positive y axis. Therefore, the electric force on the body will be in the direction of the positive y axis.
From Newton’s second law of motion we know that $F=ma$ ….. (ii),
where m is the mass and a is the acceleration of the body.
The direction of the acceleration is the direction of the force.
This means that the acceleration of the body is in the direction of the positive y axis.
From (i) and (ii) we get $F=qE=ma$.
$a=\dfrac{qE}{m}$ …. (iii).
In the case, $q={{10}^{-6}}C$, $E={{10}^{3}}V{{m}^{-1}}$ and $m=1g={{10}^{-3}}kg$.
Substitute the values in (iii).
$a=\dfrac{{{10}^{-6}}\times {{10}^{3}}}{{{10}^{-3}}}\\\ \Rightarrow a =1m{{s}^{-2}}$.
Now we know that the body is accelerating in the positive y axis direction.
Therefore, the y component of the velocity of the body will change with time. However, the x component of its velocity will remain constant since there is no force affecting its motion in that direction.

Let the x and y components of the velocity be ${{v}_{x}}$ and ${{v}_{y}}$ respectively.
At time $t=0$, ${{v}_{x}}=10m{{s}^{-1}}$ and ${{v}_{y}}=0$.
Since ${{v}_{x}}$ is constant, after time 10 seconds ${{v}_{x}}=10m{{s}^{-1}}$.
To find ${{v}_{y}}$ after time 10 seconds we will use the kinematic equation $v=u+at$ …. (iv),
where v is the body’s velocity at time t, u is its velocity at time $t=0$ and a is its acceleration.

In this case, $v={{v}_{y}}$, $u=0$, $t=10s$ and $a=1m{{s}^{-2}}$.
Substitute these values in (iv).
${{v}_{y}}=0+1(10)=10m{{s}^{-1}}$.
The resultant or net velocity of the body at this time is equal to $v=\sqrt{v_{x}^{2}+v_{y}^{2}}$.
$\Rightarrow v=\sqrt{{{10}^{2}}+{{10}^{2}}}\\\ \Rightarrow v =\sqrt{200}\\\ \therefore v =10\sqrt{2}m{{s}^{-1}}$.

Hence, the correct option is C.

Note: We can correlate the given situation with the case of a normal projectile motion of a particle projected in the presence of gravitational only. In that case, the particle will be accelerated downwards with an acceleration g, which is acceleration due to gravity.