Question

Two equal and opposite charges of 5 coulomb are kept mutually at a distance of 50 cm. The electric dipole moment of the system is:

• A. \( 5 \times 10^{-2} \) coulomb-metre
• B. \( 25 \times 10^{-2} \) coulomb-metre
• C. \( 1 \) coulomb-metre
• D. zero

Hint:

Electric dipole moment is a vector quantity directed from negative to positive charge. Formula: \[ \vec{p} = q \cdot \vec{d} \] where \( \vec{d} \) is the displacement vector from negative to positive charge.

Answer 1

The Correct Option is B

Electric dipole moment \( \vec{p} \) is given by: \[ \vec{p} = q \times 2a, \] where \( q \) is the magnitude of one charge, and \( 2a \) is the separation between the charges. Given: \[ q = 5 \, \text{C}, \quad 2a = 50 \, \text{cm} = 0.5 \, \text{m} \] So, \[ p = q \times 2a = 5 \times 0.5 = 2.5 \, \text{C} \cdot \text{m} = 25 \times 10^{-2} \, \text{C} \cdot \text{m}. \]