Two charges +3.2× 10- 19C and -3.2× 10- 19C kept 2.4 oversetoA apart form a dipole. If it is kept in uniform electric field of intensity 4× 105Vm- 1 then what will be its electrical energy in stable equilibrium.

Question

Two charges +3.2× 10- 19C and -3.2× 10- 19C kept 2.4 oversetoA apart form a dipole. If it is kept in uniform electric field of intensity 4× 105Vm- 1 then what will be its electrical energy in stable equilibrium. (A) 3× 10- 23J (B) -3× 10- 23J (C) 6× 10- 23J (D) -2× 10- 23J

Tardigrade Admin · Accepted Answer

Energy of dipole is U=- overset arrow P. overset arrow E=-PEcosθ θ =0 for stable equilibrium position ⇒ U=-PE ⇒ U =-(3.2 × 10-19 × 2.4 × 10-10)(4 × 105) =-3 × 10-23 J

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

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

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

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

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

Energy of dipole is
$U = - P \to . E \to = - PE cos θ$
$θ = 0$ for stable equilibrium position
$\Rightarrow U = - PE$
$\Rightarrow U = - (3.2 \times 1 0^{- 19} \times 2.4 \times 1 0^{- 10}) (4 \times 1 0^{5})$
$= - 3 \times 1 0^{- 23} J$

Q. Two charges +3.2×10−19C and −3.2×10−19C kept 2.4Ao apart form a dipole. If it is kept in uniform electric field of intensity 4×105Vm−1 then what will be its electrical energy in stable equilibrium.

3×10−23J

−3×10−23J

6×10−23J

−2×10−23J