Question

Two positively charged particles of charges q1 and q2 have mass m each. A uniform electric field having magnitude E exists in positive x direction as shown in figure. The given two charged particles are released from rest at t=0 as shown in figure. If position of q1 at t=2 s is given by coordinate (+2a,0) then the x coordinate of q2 at t=2s is (neglect gravitational interaction between the particles) -

Answer 1

Given that,

Two positively charged particles of charges q₁ and q₂ have mass m each.

Electric field = E

We know that,

The internal force between charges can not displace their center of mass.

We need to calculate the acceleration

Using net force on center of mass

$F=(q_{1}+q_{2})E$

Put the value into the formula

$(m+m)a=(q_{1}+q_{2})E$

$a=\dfrac{(q_{1}+q_{2})E}{2m}$

After 2 sec, displace of center of mass

We need to calculate the distance

Using equation of motion

$s=ut+\dfrac{1}{2}at^2$

Put the value into the formula

$s=0+\dfrac{1}{2}\times(\dfrac{(q_{1}+q_{2})E}{2m})\times4$

$s=\dfrac{(q_{1}+q_{2})E}{m}$

If x is displacement of the charge q₂

We need to calculate x- coordinate of q₂

Using formula for x

$s=\dfrac{m\times(2a-a)+m\times x}{2m}$

$s=\dfrac{a+x}{2}$

Put the value of s

$\dfrac{(q_{1}+q_{2})E}{m}=\dfrac{a+x}{2}$

$x=\dfrac{2(q_{1}+q_{2})E}{m}-a$

So, the coordinate of q₂ is

$x-Coordinate\ of q_{2} =-a+\dfrac{2(q_{1}+q_{2})E}{m}-a$

$x-Coordinate\ of q_{2}=\dfrac{2(q_{1}+q_{2})E}{m}-2a$

Hence, The coordinate of q₂ is $\dfrac{2(q_{1}+q_{2})E}{m}-2a$