Answer:
To find the magnitude of the electric dipole moment, we use the formula:
p = q * d
Where:
p is the magnitude of the electric dipole moment,
q is the charge of each charge in the electric dipole, and
d is the distance between the charges.
In your case, q = 3.10 C (Coulombs) and d = 8.70 μm (micrometers). However, we need to convert micrometers to meters before plugging the values into the formula. So, d = 8.70 × 10^(-6) m.
Now, we can calculate the magnitude of the electric dipole moment:
p = (3.10 C) * (8.70 × 10^(-6) m)
p = 2.697 × 10^(-5) C⋅m
Therefore, the magnitude of the electric dipole moment is 2.697 × 10^(-5) C⋅m.
Moving on to the difference between the potential energies for parallel and antiparallel dipole orientations. The potential energy of an electric dipole in an electric field is given by:
U = -p * E * cos(θ)
Where:
U is the potential energy,
p is the magnitude of the electric dipole moment,
E is the strength of the electric field, and
θ is the angle between the electric dipole moment and the electric field.
For parallel orientation, θ = 0°, and for antiparallel orientation, θ = 180°.
(a) The magnitude of the electric dipole moment is already calculated as 2.697 × 10^(-5) C⋅m.
(b) To find the difference between potential energies, we substitute the values into the formula:
U_parallel = -p * E * cos(0°) = -p * E
U_antiparallel = -p * E * cos(180°) = p * E
Taking the difference:
ΔU = U_parallel - U_antiparallel
ΔU = -p * E - (p * E)
ΔU = -2p * E
Substituting the values:
ΔU = -2 * (2.697 × 10^(-5) C⋅m) * (1290 N/C)
ΔU = -6.9686 × 10^(-2) J
Therefore, the difference between the potential energies for dipole orientations parallel and antiparallel to the electric field is approximately -6.9686 × 10^(-2) J, with a negative sign indicating a decrease in potential energy.