Question

two balls of charges q1 and q2 initially have exactly same velocity, both are subjected to equal electric fields at the same time. as a result the velocity of the first ball is reduced to half and the direction of velocity id deflected by an angle of 60 degrees. whereas the second ball gets deflected by an angle of 90.the electric fields and the initial velocity are inclined at.?

Answer 1

Answer:

Explanation:

Let v1 and v2 be the velocities of the first and second balls after the removal of the uniform electric field. By hypothesis, the angle between the velocity v1 and the initial velocity v is 60°. Therefore, the change in the momentum of the first ball is

Δp1 = q1 E = Δt = m1v sin60

Here we use the condition that v1 = v/2, which implies that the change in the momentum Δp1 of the first ball occurs in a direction perpendicular to the direction of its velocity v1

Since E || Δp1 and the direction of variation of the second ball momentum is parallel to the direction of Δp1, we obtain for the velocity of the second ball (it can easily be seen that the charges on the balls have the same sign)

V2 = vt an30 = v/√3

The corresponding change in the momentum of the second ball is

Δp2 = q2 E Δt = m2V/cos 30

Hence we obtain

q1/q2 = m1 sin60/m2/cos 30

q2/m2 = 4/3 * q1/m1 = 4/4 k1

Answer 2

Answer:

Required answer is ( 4/3)k₁

Explanation:

dp₁ = q₁Edt = m₁vsin60°

Hete we use the condition that

$V₁ = \frac{v}{2}$

Which implies that the change in the momentum dp₁ of the first ball occurs in a direction perpendicular to the direction of its velocity V₁.

Since E // dp₁ and the direction of variation of the second ball (it can easily be seen that the ccharges on the balls have the same sign)

$V₂ = Vtan30° = \frac{v}{ \sqrt{3} }$

The corresponding charge in the momentum of the second ball is

dp₂=q₂Edt =

$\frac{m₂v}{cos30°}$

Hence we obtain,

$= \frac{m₁sin°}{m₂/cos30°} ; \frac{q₂}{m₂} = \frac{4}{3} \frac{q₁}{m₁} = \frac{4}{3} \times k₁$

Where,

$k₁ = \frac{q₁}{m₁}$

Required answer is ( 4/3)k₁