Electric PE as a dipole changes its orientation in an electric field

[GodPlaysDice]

Homework Statement

The ammonia molecule (NH3) has a dipole moment of 5.0×10−30C⋅m. Ammonia molecules in the gas phase are placed in a uniform electric field E⃗ with magnitude 1.0×106N/C.
What is the change in electric potential energy when the dipole moment of a molecule changes its orientation with respect to E⃗ from parallel to perpendicular?

Homework Equations

V = U/q
ΔU = -integral([itex]qE dl[/itex])
ΔV = -integral([itex]E dl[/itex])
E(r) = -∇V(r)
I have no idea what the upside down delta is...

The Attempt at a Solution

I don't even know where to start.

 

[jackarms]

Potential energy of a dipole is given by this formula: $$U = -\stackrel{\rightarrow}{p} \cdot \stackrel{\rightarrow}{E}$$Where U is defined to be zero when the dipole is perpendicular to the field. It's derived from the formula for the torque on the dipole: [itex]\tau = \stackrel{\rightarrow}{p} \times \stackrel{\rightarrow}{E}[/itex], using the relationship [itex]U = -W = -\int^{\theta_{f}}_{\theta_{i}}\tau d\theta[/itex]