Question

Two charge qA = 2.5 × 10-7C and qB = -2.5 × 10-7C are at point A(0, 0, – 15 cm) and B (0, 0, +15 cm) respectively form a system. Determine the electric dipole moment of the system.
Solution:
Position vector of point A

Answer 1

Answer:-

The charges which are located at the given points are shown in the co-ordinate system as:-

Note:- Look into the attechment for the diagram.

At point A, total charge amount,

$qA = 2.5 \times {10}^{-7} C$

At point B, total charge amount,

$qB = -2.5 \times {10}^{-7} C$

Total charge of the system is,

$qA + qB = 2.5 \times {10}^{-7} C \times - 2.5 \times {10}^{-7} C = 0$

Distance between two charges at points A and B,

$p = qA \times d = qB \times d = 2.5 \times {10}^{- 7} \times 0.3$

$= 7.5 \times {10}^{-8} C \: m$

$p = qA \times d = qB \times d$

d = 15 + 15 = 30 cm = 0.3 m

Electric dipole moment of the system is given by,

$= 7.5 \times {10}^{-8} C \: m \: along \: positive \: z - axis$

Therefore, the electric dipole moment of the system is,

$7.5 \times {10}^{ - 8} C \: m \: along \: positive \: z-axis.$