Ncert Exercises & Solutions

Q 1.9) A system has two charges qa = 2.5 x $10^{- 7}$ $10^{- 7}$ C and qB -2.5 x $10^{- 7}$ $10^{- 7}$ 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 lof the svstem?

Both the charges can be located in a coordinate frame of reference asS
shown in the given figure.
At A, amount of charge, q

q_{A} = 2.5 \times 10^{- 7} C

$q_{A} = 2.5 \times 10^{- 7} C$
At B, amount of charge,

q_{B} = - 2.5 \times 10^{- 7} C

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

The total charge of the system,

q = q_{A} + q_{B} = 2.5 \times 10^{7} C - 2.5 \times 10^{- 7} C = 0

$q = q_{A} + q_{B} = 2.5 \times 10^{7} C - 2.5 \times 10^{- 7} C = 0$

Distance between two charges at points A and B,

d= 15 + 15 = 30 cm = 0.3 m

Electric dipole moment of the system is given by,

null m along positive Z-axis

Therefore, the electric dipole moment of the system is

7.5 \times 10^{- 8}

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

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