Question

Question : Two charges, each equal to q are kept at x = -a and x = a on the X-axis. A particle of mass m and charge $\sf q_1 = \dfrac{q}{2}$ is placed at the origin. If charge $\sf q_1$ is given a small displacement ( y << a ) along the Y-axis, the net force acting on the particle is proportional to
a. y
b. - y
c. $\sf\dfrac{1}{y}$
d. -$\sf\dfrac{1}{y}$
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Answer 1

Answer:

(A) proportional to 'y'.

Concept :

Draw a raw diagram of the system when the charge is displaced by a factor of y. At that moment, write down all the forces acting on the displaced charge. The forces are F31 (due to charge#1) , F32(due to charge#2) and mg (due to gravity).

Note : ` The force due the gravity (mg) acting on the charged particle is very small as compared to the electrostatic force (around the order of 10⁴⁰). Hence, the gravitational pull of the mass particle may be neglected. `

Now, the two forces acting of the particle, F31 and F32, needs to be added vectorially in order to obtain the net force.

To carry out this vectorial addition, I resolved the two forces into two components, one parallel to the motion of the charge (Fcos∅) and another being perpendicular to it (Fsin∅).

You must carefully notice that the two perpendicular components are equal in magnitude but opposite in direction, and hence, they nullify their effect and the net force due to these two components turns out to be zero. But, the parallel components are in same direction and hence they add up algebraically to produce a net force of 2*Fcos(∅).

Therefore, The net force, is then, only due to the parallel components, that is, 2Fcos(∅).

Calculation:

For the math, you may refer to the third attached image.

Answer 2

Topic :-

Electrostatics

Given :-

Two charges, each equal to 'q' are kept at x = -a and x = a on the X-axis.

A particle of mass 'm' and charge 'q₁ = q/2' is placed at the origin.

To Find :-

If charge 'q₁' is given a small displacement (y << a) along the Y-axis, the net force acting on the particle.

Solution :-

After displacing the particle present at origin by 'y units' along Y-axis other particles which are present on X-axis at x = a and x = -a will start applying repulsive force to it.

Magnitude of Force can be calculated using Coulomb's law.

$\mathtt{F=\dfrac{kq_1q_2}{r^2}}$

where

F = Electric Force

k = Coulomb's constant

q = Charge

r = Distance of separation between charges

Here,

r = Distance between q and q₁

From Pythagoras Theorem,

$r^2=a^2+y^2$

$r=\sqrt{a^2+y^2}$

Now,

$\mathtt{F_{net}=F\cos (90^{\circ}-\theta)+F\cos (90^{\circ}-\theta)}$

$\mathtt{F_{net}=2F\cos (90^{\circ}-\theta)}$

$\mathtt{F_{net}=2F\sin \theta}$

$(\mathtt{\because \cos (90^{\circ}-\theta)=sin \theta})$

$\mathtt{F_{net}=2\left (\dfrac{kqq_1}{a^2+y^2}\right )\left(\dfrac{y}{\sqrt{a^2+y^2}}\right)}$

$\mathtt{\left(\because F=\dfrac{kqq_1}{a^2+y^2}\right )}$

$\mathtt{\left( \because\sin\theta=\dfrac{y}{\sqrt{a^2+y^2}} \right)}$

$\mathtt{F_{net}=2\left (\dfrac{kq^2}{2(a^2+y^2)}\right )\left(\dfrac{y}{\sqrt{a^2+y^2}}\right)}$

$\mathtt{\left(\because q_1=\dfrac{q}{2}\right )}$

$\mathtt{F_{net}=\not{2}\left (\dfrac{kq^2y}{\not{2}(a^2+y^2)^{3/2}}\right )}$

$\mathtt{F_{net}=\dfrac{kq^2y}{(a^2+y^2)^{3/2}}}$

$\mathtt{F_{net}=\dfrac{kq^2y}{(a^2)^{3/2}}}$

$(\because y<<a)$

$\mathtt{F_{net}=\dfrac{kq^2y}{a^3}}$

We can clearly see that,

$\mathtt{F_{net} \propto y}$

Answer :-

The net force acting on the particle is proportional to y. Hence, option (a) is correct option.

Note :

For better understanding refer to the attachment.