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, 1.20 mm) and the negative charge is at the point (1.70 mm, −1.30 mm).
Required:
a. Find the electric dipole moment of the object.
b. The object is placed in an electric field E = (7.80 103 î − 4.90 103 ĵ). 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,
Required:
a. Find the electric dipole moment of the object.
b. The object is placed in an electric field E = (7.80 103 î − 4.90 103 ĵ). 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,
1 answer:
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Answer:
C = 17 i^ - 7 j^ + 16 k^ , | C| = 24.37
Explanation:
To work the vactor component method, we add the sum in each axis
C = A + B = (Aₓ + Bₓ) i ^ + ( +
) i ^ + (
+
) k ^
Cₓ = 12+ 5 = 17
= -37 +30 = -7
= 58 -42 = 16
Resulting vector
C = 17 i ^ - 7j ^ + 16k ^
The mangitude of the vector is
| C | = √ c²
| C | = √( 17² + 7² + 16²)
| C| = 24.37