Torque on a dipole in an electric field

[castrodisastro]

Homework Statement

An electric dipole has opposite charges of 6.55⋅10⁻¹⁵C separated by a distance of 0.65mm. It is oriented at 54.0° with respect to a uniform electric field of magnitude 10.0⋅10³N/C. Determine the magnitude of the torque exerted on the dipole by the electric field.

q₊=6.55*10^-15C
q_-=-6.55*10^-15C
r=0.00065m
\vec{E}=10,000N/C
Θ=54.0°

Homework Equations

p=qd
\tau=p\times\vec{E}

The Attempt at a Solution

The angle of the dipole is 54.0° with respect to the uniform electric field. Using trig I can see that the torque can be calculated by the cross product of the dipole moment and the magnitude of the electric field multiplied by sin(Θ).

\tau=p\times\vec{E}

The dipole moment is given by charge times the distance(r)

p=qd

p=(6.55*10^-15C)(0.00065m)=4.26*10^-18C*m

If we plug in that value into our torque equation...

\tau=p\times\vec{E}sin(Θ)

\tau=(4.26*10^-18C*m)*(10,000N/C)*sin(54°)

\tau=3.44*10^-14N*m

Which is not correct according to my online homework. This calculation seems simple...too simple. Am I missing something?

Also, if the torque results from a cross product operation, which is an operation between vectors, then in this case I just multiply the magnitude of the dipole moment and the magnitude of the electric field because we are not working with vectors in this problem?

As always, any help is appreciated. Thanks in advance.

 

[haruspex]

\tau=3.44*10^-14N*m

I get the same.

Also, if the torque results from a cross product operation, which is an operation between vectors, then in this case I just multiply the magnitude of the dipole moment and the magnitude of the electric field because we are not working with vectors in this problem?

No, you must include the sin(). Is that what you meant?