Get Answers to all your Questions

header-bg qa

. Two charges +3.2*10^{-19} and-3.2*10^{-19} C  placed at 2.4A apart form an electric dipole. It is placed in a uniform electric field of intensity 4*10^5 volt/m . The electric dipole moment is

  • Option 1)

    15.36*10^{-29 } coulomb\times m

  • Option 2)

    15.36*10^{-19 } coulomb\times m

  • Option 3)

    7.68*10^{-29 } coulomb\times m

  • Option 4)

    7.68*10^{-19 } coulomb\times m

 

Answers (1)

best_answer

As we have learned

Dipole moment -

\left ( \vec{P} \right )=q\left ( \vec{2l} \right )

S.I unit - C-m or Debye

1\, Debye=3.3\times 10^{-30}c-m

 

 

- wherein

2l\rightarrow dipole\, length

 

 

 Dipole moment p = q (2l)

=3.2\times 10^{-19}\times (2.4\times 10^{-10})= 7.68\times 10^{-29}Cm

 


Option 1)

15.36*10^{-29 } coulomb\times m

Option 2)

15.36*10^{-19 } coulomb\times m

Option 3)

7.68*10^{-29 } coulomb\times m

Option 4)

7.68*10^{-19 } coulomb\times m

Posted by

Plabita

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE