1)a molecule of a substance has permanent electric dipole moment equal to 10to the power (-29) Cm. a mole of this substance is polarised at low temperature by applying a strong electrostatic field of magnityde 10 to the power 6 V/m.).th direction of this field is suddenly changed by an angle of 60degrees. estimate the heat released by the substance in aligning its dipoles along the new direction of the field. (for simplicity assume100%polarisation of sample).

2)two identical balls of charges q1 nd q2 initially have equal velocity of the same magnitude nd direction. after a uniform electric field is for some time ,the direction of the first ball changes by 60degrees and the magnitude is reduced to half. the direction of the velocity of second ball changes therby by 90degres. by what factor will the velocity of the second ball change?

[I will get the answers today. dont know it rte now]

I know the ans of 2nd question man.

wait i m giving the ans.

by conservation of momentum

v₁/v₂=tan

v₁/v₂=tan 60=root 3

v₂=v₁/root 3

hope u got it.

yes i agree with rhd for ist ques.

rhd i m also from bhilai

i, j are unit vectors along x, y axes

let initial velocity of both charges be v i

Impulse = change in momentum

q1Et = m[(v/2)(cos60 i + sin60 j) - v i] ........(let its velcity nw makes a positive angle 60 with x-axis )

or Et/m = 1/q1[(-3v/4 i + rt(3)v/4 j].............................(1)

Let for q2, final velocity = v2 (vector)

so, Et/m = 1/q2 [ v2 - v i]..........................................(2)

1/q1[-3v/4 i + rt(3)v/4 j] = 1/q2[v2-v1]

v2 = (v - 3q2v/4q1) i + rt(3)q2v/4q1 j

Since, its direction has changed by 90deg, i component =0

v - 3q2v/4q1 = 0

q2/q1 = 4/3

So, v2 = rt(3)v/4 * (q2/q1) j = rt(3)v/4 * 4/3 = v/rt(3)

mag of  vel changes by a factor of 1/rt(3)