A mal, rigid object carries positive and negative 2.00 nC charges. St is onented

Question

A mal, rigid object carries positive and negative 2.00 nC charges. St is onented so that the positive chage has coondinates (-1.20mm, 1.00 mm) and the negative charge is at the point (1.90 mm, -1.30 mm (a) Find the electric dipole moment of the object (b) The obiect s placed in an.lectrg held E.(7.so ×1031.4.90 X 1033)N/C. Fird torqu" scting on th" obiect. Your respoes..wenn 10% of the correct valus. This may be du° to roundolfarroc er you could have matake "your calculation, Carry 0st all intemeiatereadtst.t (c) Find th·potential energy of the object-field system when the object 1" in this orientate response difers from the correct answer by more than 10%. Dadle check your tals-ins.) (d) Amming the orientaten of the object can charge, fnd the dumerena betveen the marmum .in.- um poteta-ege. efth Noed Help?

Explanation / Answer

Part A:

Vector_P = Vector_2a*q

Vector_2a = (-1.2-1.9)i + {1-(-1.3)}j

Vector_2a = - 3.1 i + 2.3 j

Vector_P = 2nC*(- 3.1 x 10-3 i + 2.3 x 10-3 j)m

Vector_P = 2 x 10-9*(- 3.1 x 10-3i + 2.3 x 10-3 j)

Vector_P = -6.2 x 10-12 C-m i + 4.6 x 10-12C-m j

Part B

Vector_t = Vector_p X Vector_E

                  |i                   j                   k|

Vector_t = |-6.2 x 10-12    4.6 x 10-12       0|

                  |7800           -4900          0|

Vector_t = (30380 x 10-12 – 35880 x 10-12 )k

Vector_t = -5.5 x 10-9 N-m k

Part C

P.E. = -Vector_p . Vector_E

P.E. = -(-6.2 x 10-12 i + 4.6 x 10-12 j).(7800i – 4900j)

P.E. = 1.09 x 10-7 j

Part D:

P.Emax = |Vecor_P| |Vector_E|

P.Emax = [sqrt(6.22+4.62)] x 10-12*[78002+49002]

P.Emax = 7.72 x10-12*9211.4059

P.Emax = 7.11 x10-8 J

Delta P.E = 2*P.Emax

Delta P.E = 1.4224 x10-7 J

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