How much work is required to turn an electric dipole 180 degrees in a uniform electric field of magnitude 37.1 N/C? The electric dipole moment is [tex] p = 3.45 \times 10^{-25} \, \text{C} \cdot \text{m} [/tex] and the initial angle is 20.3 degrees.

What is the correct amount of work needed?

A) 37.1 N/C
B) [tex] 3.45 \times 10^{-25} \, \text{C} \cdot \text{m} [/tex]
C) 180 degrees
D) 20.3 degrees

The work required to turn an electric dipole 180 degrees in a uniform electric field can be calculated using the equation W = pE(1 - cosθ), where p is the dipole moment, E is the magnitude of the electric field, and θ is the change in angle. The correct multiple-choice option is (b).

Explanation:

To calculate the amount of work required to turn an electric dipole 180 degrees in a uniform electric field, you would use the potential energy equation for an electric dipole in an electric field: W = pE(1 - cosθ). Here, W is the work done, p is the dipole moment, E is the magnitude of the electric field, and θ is the angle through which the dipole is turned.

In this case, the initial angle is 20.3 degrees, and we are turning the dipole 180 degrees to an angle of 180 + 20.3 degrees. The potential energy change, which is equal to the work done, will be W = pE(1 - cos(180 + 20.3 degrees)). We need to calculate this value using the given electric field E = 37.1 N/C and the given dipole moment p = 3.45 x 10-25 C*m.

Since cos(180 + θ) is equal to -cos(θ), the equation for work simplifies to W = pE(1 - (-cos(20.3 degrees))). Plugging the values into the equation will give us the answer in joules, which is the unit of work.

The correct multiple-choice option is (b).