a) Elctric dipole moment (p) = q x d
p = 4.20 x 10-9 C x 4 x 10-3 m = 1.68 x 10-11 C m
b) q1 to q2
c) Torque = p x E = p E sin(36.6o)
== 7.4 x 10 -9 = 1.68 x 10 -11 x E x 0.596
E = 7.39 x 10 2 N/C
SOLUTION :
a.
q = magnitude of the couple formed = 4.20 nC
Electric Di-pole moment (magnitude) of given charges of opposite but equal magnitude :
= magnitude of charge in C * distance in m
= 4.20 * 10^(- 9) * 4 * 10^(-3)
= 16.80 * 10^(-12)
= 1.68 * 10^(-11) (C-m) (ANSWER).
b.
Direction of the electric dipole moment is from negative charge to positive charge.
So, this direction is from q1 to q2. (ANSWER).
c.
Torque magnitude
= Polar moment magnitude * Electric field E magnitude * sin (angle between them)
=> 7.40 * 10(-9) = 1.68*10(-11) * E (magnitude) * sin(36.6)
=> E (magnitude) = 7.40 * 10^(-9) / (1.68 * 10^(-11) * sin(36.6))
=> E (magnitude) = 738.78 N/C (ANSWER).
SOLUTION :
a.
q = magnitude of the couple formed = 4.20 nC
Electric Di-pole moment (magnitude) of given charges of opposite but equal magnitude :
= magnitude of charge in C * distance in m
= 4.20 * 10^(- 9) * 4 * 10^(-3)
= 16.80 * 10^(-12)
= 1.68 * 10^(-11) (C-m) (ANSWER).
b.
Direction of the electric dipole moment is from negative charge to positive charge.
So, this direction is from q1 to q2. (ANSWER).
c.
Torque magnitude
= Polar moment magnitude * Electric field E magnitude * sin (angle between them)
=> 7.40 * 10(-9) = 1.68*10(-11) * E (magnitude) * sin(36.6)
=> E (magnitude) = 7.40 * 10^(-9) / (1.68 * 10^(-11) * sin(36.6))
=> E (magnitude) = 738.78 N/C (ANSWER).
Exercise 21.57 ia Exercise 21.57 Point charges h =-4.20 nC and g2 4.20 nC are separated...
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