$A$ system has two charges $q_{A} = 2.5 \times 10^{-7} \; C$ and $q_{B} = -2.5 \times 10^{-7} \; C$ located at points $A: (0, 0, -15 \; cm)$ and $B: (0, 0, +15 \; cm)$,respectively. What are the total charge and electric dipole moment of the system?
(N/A) The total charge of the system is the algebraic sum of the individual charges:
$q_{total} = q_{A} + q_{B} = 2.5 \times 10^{-7} \; C + (-2.5 \times 10^{-7} \; C) = 0 \; C$
The distance between the two charges is the distance between points $A(0, 0, -15 \; cm)$ and $B(0, 0, 15 \; cm)$:
$d = 15 \; cm - (-15 \; cm) = 30 \; cm = 0.3 \; m$
The electric dipole moment $p$ is defined as the product of the magnitude of one of the charges and the distance between them:
$p = |q| \times d = (2.5 \times 10^{-7} \; C) \times (0.3 \; m) = 7.5 \times 10^{-8} \; C \cdot m$
Since the dipole moment is directed from the negative charge to the positive charge,and the negative charge is at $z = -15 \; cm$ and the positive charge is at $z = +15 \; cm$,the direction of the dipole moment is along the positive $z$-axis.
$q_{total} = q_{A} + q_{B} = 2.5 \times 10^{-7} \; C + (-2.5 \times 10^{-7} \; C) = 0 \; C$
The distance between the two charges is the distance between points $A(0, 0, -15 \; cm)$ and $B(0, 0, 15 \; cm)$:
$d = 15 \; cm - (-15 \; cm) = 30 \; cm = 0.3 \; m$
The electric dipole moment $p$ is defined as the product of the magnitude of one of the charges and the distance between them:
$p = |q| \times d = (2.5 \times 10^{-7} \; C) \times (0.3 \; m) = 7.5 \times 10^{-8} \; C \cdot m$
Since the dipole moment is directed from the negative charge to the positive charge,and the negative charge is at $z = -15 \; cm$ and the positive charge is at $z = +15 \; cm$,the direction of the dipole moment is along the positive $z$-axis.
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