50. A small, rigid object carries positive and negative 3.50-nC charges. It is o

Question

50. A small, rigid object carries positive and negative 3.50-nC charges. It is oriented so that the positive charge has coordinates (-1.20 mm, 1.10 mm) and the negative charge is at the point (1.40 mm, -1.30 mm). (a) Find the electric dipole moment of the object. The object is placed in an electric field E-(7.80 × 101 4.90 × 10') N/C (b) Find the torque acting on the object. (c) Find the potential energy of the object-field system when the object is in thi orientation. (d) Assum- the difference between the maximum and minimum potential energies of the system.

Explanation / Answer

given
charge q = 3.5*10^-9 C
location of +q = (-1.2, 1.1) mm
location of -q = (1.4, -1.3) mm
so a vector joining -q to +q is given by d
d = (-1.2-1.4,1.1+1.3) = (-2.6, 2.4)
let i and j be unit vectors along x and y directions
then
d = -2.6i + 2.4j

a. hence dipole moment p = qd = 3.5*10^-9(-2.6i + 2.4j)
   p = [-9.1i + 8.4j]*10^-12 Cm

b. given electric field, E = 7.8*10^3 i - 4.9*10^3 j N/C
   torque on object = T
   T = p x E = ([-9.1i + 8.4j]*10^-12 ) x (7.8*10^3 i - 4.9*10^3 j)
   T = (44.59k - 65.52k)*10^-9 = -20.93*10^-9 k Nm

c. electric potential energy of a dipole in umiform electric field is given by
   U = -p.E
   U = -([-9.1i + 8.4j]*10^-12 ) . (7.8*10^3 i - 4.9*10^3 j)
   U = 112.14*10^-9 J

d. Electric field magnitude |E| = sqroot(7.8^2 + 4.9^2)*10^3 = 9.2114*10^3 N/C
   dipole moment magnitude |p| = sqroot(9.1^2 + 8.4^2)*10^-12 = 12.384*10^-12 Cm
   hence maximum potential energy = |p||E|
   minimum potential energy = -|p||E|
   difference between these two energies = 2|p||E| = 2*9.2114*12.384*10^-9 J = 228.1479552*10^-9 J

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