Question

Two charges $+3.2\times 10^{- 19}C$ and $-3.2\times 10^{- 19}C$ kept $2.4\overset{o}{A}$ apart form a dipole. If it is kept in uniform electric field of intensity $4\times 10^{5}Vm^{- 1}$ then what will be its electrical energy in stable equilibrium.

• A. $3\times 10^{- 23}J$
• B. $-3\times 10^{- 23}J$
• C. $6\times 10^{- 23}J$
• D. $-2\times 10^{- 23}J$

Answer 1

Answer: (B)

Energy of dipole is
$U=-\overset{ \rightarrow }{P}.\overset{ \rightarrow }{E}=-PEcos\theta $
$\theta =0$ for stable equilibrium position
$\Rightarrow U=-PE$
$\Rightarrow U =-\left(3.2 \times 10^{-19} \times 2.4 \times 10^{-10}\right)\left(4 \times 10^5\right)$
$=-3 \times 10^{-23} J$