(a) Derive an expression for the torque experienced by an electric dipole kept in a uniform electric field.
(b) Calculate the work done to dissociate the system of three charges placed on the vertices of a triangle as shown.
Here `q= 1.6 xx 10^(-10)C`.

Explanation

(a) Dipole in a Uniform External Field:
Consider an electric dipole consisting of two equal and opposite point charges separated by a small distance AB =2a having dipole moment
`|vec(P)|=q(2a)`
Let the dipole held in a uniform external electric field `vec(E)` at an angle `theta`.
`:.` Force on charge (+q) =+q E along the direction of `vec(E)`
Force on charge (-q) = - q `vec(E)` along the opp. direction of `vec(E)`
`:.` The net forces on hte dipole is zero since `vec(E)` is uniform.
These forces being equal, unlike and parallel from a couple, which rotates the dipole in clockwise direction.
`:.` Magnitude of torque = force`xx` arm of couple
`tau=E.AC = q E A B sin theta = (q E) 2a sin theta`
`tau = q (2a ) E sin theta`
`tau = PE sin theta [:' P=q(2a)]`
`:. vec(tau) = vec(P)xx vec(E)`
The direction of `vec(tau)` is given by right hand screw rule and is normal to `vec(P) and vec(E)`.
(b) Let `q_(1)=q, q_(2)=-4q and q_(3)=+2q`
and `r=10 cm = 0.10 m `
Total work done = `(1)/(4 pi in_(0)) (q_(1)q_(2))/(r) + (1)/(4 pi in_(0)) (q_(2)q_(3))/(r) + (1)/(4 pi in_(0)) (q_(3)q_(1))/(r)`
`=(1)/(4pi in_(0)r)(q_(1)q_(2)+q_(2)q_(3)+q_(3)q_(1))=(1)/(4pi in_(0)r)[(q)(-4q)+(-4q)(+2q)+(+2q)(q)]`
`=(1)/(4pi in_(0)r) [-4q^(2)-8q^(2)+2q^(2)]=(1)/(4pi in_(0)r) (-10q^(2))`
`=9xx10^(9)xx(-10xx(1.6xx10^(-10))^(2))/(0.10)=(-9xx10^(9)xx10xx2.56xx10^(-20)xx100)/(10)`
`=-23.04xx10^(-9)=-2.304xx10^(-6)J`.