A small, rigid object carries positive and negative 4.00 nC charges. It is orien

Question

A small, rigid object carries positive and negative 4.00 nC charges. It is oriented so that the positive charge has coordinates (?1.20 mm, 1.40 mm) and the negative charge is at the point (1.70 mm, ?1.30 mm).

(b) The object is placed in an electric field

Find the torque acting on the object.

(c) Find the potential energy of the object–field system when the object is in this orientation.

(d) Assuming the orientation of the object can change, find the difference between the maximum and the minimum potential energies of the system.

A small, rigid object carries positive and negative 4.00 nC charges. It is oriented so that the positive charge has coordinates (-1.20 mm, 1.40 mm) and the negative charge is at the point (1.70 mm,-1.30 mm) (a) Find the electric dipole moment of the object. ( b) The object is placed in an electric field E = (7.80 x 103-4.90 x 10-j) N/C. Find the torque acting on the object . (c) Find the potential energy of the object-field system when the object is in this orientation (d) Assuming the orientation of the object can change, find the difference between the maximum and the minimum potential energies of the system

Explanation / Answer

The electric dipole moment is a vector

p = SUM of (q_i*r_i)

where q_i is the charge and r_i is the position vector of the i-th charge. Here r_1 = (-1.2, 1.40, 0) mm and r_2 = (1.7, -1.3, 0) mm and q_1 = + 4 nC and q_2 = - 4 nC

So the dipole moment vector is the sum

p = 4*(-1.2, 1.40, 0) - 4*(1.7, -1.3, 0) in units of mm nC

p = (-4.8-6.8, 5.6+5.2, 0) mm nC

p = (- 11.6, 10.8, 0) mm nC

px = -11.6 mm nC

py = 10.8 mm nC

pz = 0.0 mm nC

b)

Torque is given by

Torque = p x E    [cross product of vector p and vector E]

vector p = (- 11.6 i + 10.8 j + 0 k ) mm nC

vector E = (7.8*10^3 i - 4.9*10^3 j) N/C

Torque = [11.6*4.9*10^3 (ixj) + 10.8*7.8*10^3 (jxi)] = - 27400 k (mm nC) N/C = - 27400*10^(-12) N.m

c) The potential energy can be found using the dot product

U = - p . E

U = - [- 11.6*7.8*10^3 - 10.8*4.9*10^3] = 143400 J

d) The extremes of potential energy occur when the field and the dipole are parallel and antiparallel. So the difference would be 2|p| |E|.

difference between potential energy = 2*sqrt(11.6^2 + 10.8^2)*sqrt((7.8*10^3)^2 + (4.9*10^3)^2) = 2*15.85*9211.4 = 292001.38 J

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