A single water molecule were oriented such that its dipole moment (magnitude 6.186 * 10 - 30 Cm is along a z axis. How much torque does an 8500 N/C electric field exert on the dipole if the field lies in x z - plane, 42 ∘ above the x - axis?

Question

Jonas

Physics

5 January, 02:17

A single water molecule were oriented such that its dipole moment (magnitude 6.186 * 10 - 30 Cm is along a z axis. How much torque does an 8500 N/C electric field exert on the dipole if the field lies in x z - plane, 42 ∘ above the x - axis?

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Answers (1)

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Tanner Nixon · Answer 1

Torque=6.261*10^-43Nm

Explanation:

Torque is given by:

Torque=dqEsin (phi)

Where E is a constant=1.6022*10^-19

d = distance = 6.186*10^-30cm

Changing to metres=6.186*10^-28m

q = the charge=8500NC

Phi = 42°

Torque = (6.186*10^-28) * 8500 * (1.6022*10^-19) sin42°

Torque = 8.425*10^-43 * 0.7431

Torque = 6.261 * 10^-43