Question

b) An electric dipole in a uniform electric field of magnitude 5.0×105 N/C that is directed parallel to the plane of the figure. The charges are
±1.6×10-19 C; both lie in the plane and are separated by 0.125 nm =
0.125 ×10-9 m. Find (i) the magnitude and direction of the electric dipole moment; (ii) the magnitude and direction of the torque.

Answer 1

The Charges Are Plusminus 1.6 X 10-19 C; Both Lie In The Plane And Are Separated ... Find The Net Force Exerted By The Field On The Dipole; The Magnitude And ... field with magnitude 5.0 x 105 N/C directed parallel to the plane of the figure. ... Figure 1.0 (a) an electric dipole (b) Direction of the electric dipole moment, ...

Answer 2

Given:

|E| = 5.0×10^5 N/C

q1 = +1.6×10^-19 C

q2 = -1.6×10^-19 C

|q| = 1.6×10^-19 C

d = 0.125×10^-9 m

To Find :

i) The magnitude and the direction of the electric dipole moment(P).

ii)  The magnitude and the direction of the torque(T).

Solution:

i)  For the magnitude of electric dipole moment, we know

    |P| = |q| × |d|

         =  1.6×10^-19 × 0.125×10^-9

         =  2×10^-27 Cm

The direction of the dipole moment is always from negative to positive charge.

ii) For the magnitude of the torque

    |T| = |P| × |E| Sinθ  , ( θ = 90 , The dipole is parallel to plane of figure)

         = 2 × 10^-27 × 5 × 10^5

         = 1 × 10^-21 NC^2/m

The direction of the torque is clockwise.

Hence, the magnitude of the dipole moment is 2×10^-27 Cm

Hence, the magnitude of Torque is 1 × 10^-21 NC^2/m

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