An electric dipole, consisting of two oppositecharges of 2x10-6 C each...
Charges (q) = 2 × 10
–6 C, Distance (d)
= 3 cm = 3 × 10
–2 m and electric field (E)

View all questions of this testAn electric dipole, consisting of two oppositecharges of 2x10-6 C each...
An electric dipole, consisting of two oppositecharges of 2x10-6 C each...
Microcoulombs separated by a distance of 1 cm, is placed in a uniform electric field of magnitude 10 N/C, directed along the positive x-axis.
(a) Find the magnitude of the electric force on the positive charge of the dipole.
(b) Find the magnitude of the torque on the dipole.
(c) In what direction does the torque tend to rotate the dipole?
(d) If the dipole initially makes an angle of 30 degrees with the x-axis, what is the magnitude of the torque that tends to align the dipole with the field direction?
(a) The electric force on the positive charge of the dipole is given by F = qE, where q is the charge on the positive charge and E is the electric field strength. Thus, F = (2 microC)(10 N/C) = 20 microN.
(b) The torque on the dipole is given by τ = p x E, where p is the dipole moment and x is the vector cross product. Since the dipole moment points from the negative charge to the positive charge, and the electric field is along the positive x-axis, we have p x E pointing in the negative y-axis direction. Thus, τ = pE sinθ = (2 microC)(0.01 m)(10 N/C) sin(90 degrees) = 0.2 microN·m.
(c) The torque tends to rotate the dipole so that its moment points in the same direction as the electric field, i.e. along the positive x-axis.
(d) The torque that tends to align the dipole with the field direction is given by τ = pE sinθ, where θ is the angle between the dipole moment and the electric field direction. Since the dipole initially makes an angle of 30 degrees with the x-axis, we have sinθ = sin(30 degrees) = 0.5. Thus, τ = (2 microC)(0.01 m)(10 N/C)(0.5) = 0.1 microN·m.