Two charges +3.2 × 10-19 C and -3.2 × 10-19 C kept 2.4 mathringA apart form a dipole. If it is kept in uniform electric field of intensity 4 × 105 volt / m, then what will be its electrical energy in equilibrium?

Question

Two charges +3.2 × 10-19 C and -3.2 × 10-19 C kept 2.4 mathringA apart form a dipole. If it is kept in uniform electric field of intensity 4 × 105 volt / m, then what will be its electrical energy in equilibrium? (A) +3 × 10-23 J (B) -3 × 10-23 J (C) -6 × 10-23 J (D) -2 × 10-23 J

Tardigrade Admin · Accepted Answer

Potential energy of electric dipole U=-p E cos θ=-(q × 2 l) E cos θ U=-(3.2 × 10-19 × 2.4 × 10-10) 4 × 105 cos θ U=-3 × 10-23 J (approx.)

Q. Two charges $+ 3.2 \times 1 0^{- 19} C$ and $- 3.2 \times 1 0^{- 19} C$ kept $2.4 \overset{˚}{A}$ apart form a dipole. If it is kept in uniform electric field of intensity $4 \times 1 0^{5} v o lt / m$ , then what will be its electrical energy in equilibrium?

$+ 3 \times 1 0^{- 23} J$

$- 3 \times 1 0^{- 23} J$

$- 6 \times 1 0^{- 23} J$

$- 2 \times 1 0^{- 23} J$

Potential energy of electric dipole
$U = - pE cos θ = - (q \times 2 l) E cos θ$
$U = - (3.2 \times 1 0^{- 19} \times 2.4 \times 1 0^{- 10}) 4 \times 1 0^{5} cos θ$
$U = - 3 \times 1 0^{- 23} J$ (approx.)

Q. Two charges +3.2×10−19C and −3.2×10−19C kept 2.4A˚ apart form a dipole. If it is kept in uniform electric field of intensity 4×105volt/m, then what will be its electrical energy in equilibrium?

+3×10−23J

−3×10−23J

−6×10−23J

−2×10−23J