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Question: Two charges \(+ 3.2 \times 10^{- 19}C\) and \(- 3.2 \times 10^{- 9}C\) kept 2.4 Å apart forms a dipo...

Two charges +3.2×10−19C+ 3.2 \times 10^{- 19}C and −3.2×10−9C- 3.2 \times 10^{- 9}C kept 2.4 Å apart forms a dipole. If it is kept in uniform electric field of intensity 4×105volt⥂/⥂m4 \times 10^{5}volt ⥂ / ⥂ m then what will be its electrical energy in equilibrium

A

+3×10−23J+ 3 \times 10^{- 23}J

B

−3×10−23J- 3 \times 10^{- 23}J

C

−6×10−23J- 6 \times 10^{- 23}J

D

−2×10−23J- 2 \times 10^{- 23}J

Answer

−3×10−23J- 3 \times 10^{- 23}J

Explanation

Solution

Potential energy of electric dipole

U=−pEcos⁡θU = - pE\cos\theta =−(q×2l)Ecos⁡θ= - (q \times 2l)E\cos\theta

U=−(3.2×10−19×2.4×10−10)4×105cos⁡θU = - (3.2 \times 10^{- 19} \times 2.4 \times 10^{- 10})4 \times 10^{5}\cos\theta

U=−3×10−23U = - 3 \times 10^{- 23} (approx.)