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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?

Respuesta :

Answer:

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