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Question: Two charges\(+ 3.2 \times 10^{- 19}\) and \(- 3.2 \times 10^{- 19}C\) placed at \(2.4Å\) apart form ...

Two charges+3.2×10−19+ 3.2 \times 10^{- 19} and −3.2×10−19C- 3.2 \times 10^{- 19}C placed at 2.4A˚2.4Å apart form an electric dipole. It is placed in a uniform electric field of intensity4×105volt/m4 \times 10^{5}volt/m. The electric dipole moment is

A

15.36×10−29 6mucoulomb×m15.36 \times 10^{- 29}\mspace{6mu} coulomb \times m

B

15.36×10−19 6mucoulomb×m15.36 \times 10^{- 19}\mspace{6mu} coulomb \times m

C

7.68×10−29 6mucoulomb×m7.68 \times 10^{- 29}\mspace{6mu} coulomb \times m

D

7.68×10−19 6mucoulomb×m7.68 \times 10^{- 19}\mspace{6mu} coulomb \times m

Answer

7.68×10−29 6mucoulomb×m7.68 \times 10^{- 29}\mspace{6mu} coulomb \times m

Explanation

Solution

Dipole moment p = q (2l)

=3.2×10−19×(2.4×10−10)=7.68×10−29C−m= 3.2 \times 10^{- 19} \times (2.4 \times 10^{- 10}) = 7.68 \times 10^{- 29}C - m