Two charges +3.2 × 10-19 and -3.2 × 10-19 C placed at 2.4 mathringA apart form an electric dipole. It is placed in a uniform electric field of intensity 4 × 105 V / m. The electric dipole moment is

Question

Two charges +3.2 × 10-19 and -3.2 × 10-19 C placed at 2.4 mathringA apart form an electric dipole. It is placed in a uniform electric field of intensity 4 × 105 V / m. The electric dipole moment is (A) 15.36 × 10-29 Cm (B) 15.36 × 10-19 Cm (C) 7.68 × 10-29 Cm (D) 7.68 × 10-19 Cm

Tardigrade Admin · Accepted Answer

Dipole moment, p=q(2 l) =3.2 × 10-19 ×(2.4 × 10-10) =7.68 × 10-29 C - m

Q. Two charges $+ 3.2 \times 1 0^{- 19}$ and $- 3.2 \times 1 0^{- 19} C$ placed at $2.4 \overset{˚}{A}$ apart form an electric dipole. It is placed in a uniform electric field of intensity $4 \times 1 0^{5} V / m$ . The electric dipole moment is

$15.36 \times 1 0^{- 29} C m$

$15.36 \times 1 0^{- 19} C m$

$7.68 \times 1 0^{- 29} C m$

$7.68 \times 1 0^{- 19} C m$

Dipole moment, $p = q (2 l)$
$= 3.2 \times 1 0^{- 19} \times (2.4 \times 1 0^{- 10})$
$= 7.68 \times 1 0^{- 29} C - m$

Q. Two charges +3.2×10−19 and −3.2×10−19C placed at 2.4A˚ apart form an electric dipole. It is placed in a uniform electric field of intensity 4×105V/m. The electric dipole moment is

15.36×10−29Cm

15.36×10−19Cm

7.68×10−29Cm

7.68×10−19Cm