A small, rigid object carries positive and negative 3.00 nC charges. It is orien

Question

A small, rigid object carries positive and negative 3.00 nC charges. It is oriented so that the positive charge has coordinates (1.20 mm, 1.60 mm) and the negative charge is at the point (1.70 mm, 1.30 mm).

Find the torque acting on the object.
N · m ---Select--- i j k

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

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

Explanation / Answer

charge q=3nC

position of +ve charge,(-1.2 mm, 1.6mm)

and

position of -ve charge, (1.7mm, -1.3 mm)

displacement vector, r=(-1.2-1.7)i+(1.6-(-1.3)) j

r=(-2.9)i+(2.9)j in mm

a)

dipole moment, P=q*r

P=3*10^-9*((-2.9)i+(2.9)j)*10^-3 C.m

P=[(-8.7*10^-12)i+(8.7*10^-12)j] C.m

b)

if electric field E=[7.8*10^3)i+(-4.9*10^3)j] N/C

torque, T=PXE

T=[(-8.7*10^-12)i+(8.7*10^-12)j]X[7.8*10^3)i+(-4.9*10^3)j]

T=(42.63*10^-9)k-(67.86*10^-9)k

T=(-25.23*10^-9)k N.m

c)

potential energy,

U=-P.E

U=[(-8.7*10^-12)i+(8.7*10^-12)j].[7.8*10^3)i+(-4.9*10^3)j]

U=(-67.86*10^-9)-(42.63*10^-9)

U=(-25.23*10^-9)k N.m

U=110.5*10^-9 J

U=110.5 nJ

d)

magnitude of P is, P=sqrt([(-8.7*10^-12)^2+(8.7*10^-12)^2]

P=1.23*10^-11 C.m

and

magnitude of E=sqrt((7.8*10^3)^2+(-4.9*10^3)^2)

E=9.21*10^3 N/C

Umax=P*E

Umax=1.23*10^-11*9.21*10^3

Umax=113.3 nJ

and

Umin=115.5 nJ

difference d=(115.5) -(-110.5)

d=226 nJ

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