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  • How to solve this problem- - Electrostatics - JEE Main-5

Two opposite and equal charges 4\times 10^{-8} when placed 2\times 10^{-2} away, form a dipole. If this dipole is placed in an external electric field 4\times 10^{-8} , the value of maximum torque and the work done in rotating it through 180 \degree  will be

  • Option 1)

    64\times 10^{-4}Nm ..and..64\times 10^{-4}J

  • Option 2)

    32\times 10^{-4}Nm ..and..32\times 10^{-4}J

  • Option 3)

    64\times 10^{-4}Nm ..and..32\times 10^{-4}J

  • Option 4)

    32\times 10^{-4}Nm ..and..64\times 10^{-4}J

 

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 = 4 ´ 10–8´ 2 ´ 10–4 = 8 ´ 10–12m

Maximum torque = pE = 8 ´ 10–12´ 4 ´ 108

= 32 ´ 10–4Nm

Work done in rotating through 180o = 2pE

= 2 ´ 32 ´ 10–4 = 64 ´ 10–4J

 


Option 1)

64\times 10^{-4}Nm ..and..64\times 10^{-4}J

Option 2)

32\times 10^{-4}Nm ..and..32\times 10^{-4}J

Option 3)

64\times 10^{-4}Nm ..and..32\times 10^{-4}J

Option 4)

32\times 10^{-4}Nm ..and..64\times 10^{-4}J

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