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Give answer! - Electrostatics - JEE Main-9

A system of three charges are placed as shown in the figure :

If D> > d, the potential energy of the system is best given by :

  • Option 1)

      \frac{1}{4\pi \epsilon _{0}}\left [ -\frac{q^{2}}{d}-\frac{q\; Q\; d}{2D^{2}} \right ]

  • Option 2)

       \frac{1}{4\pi \epsilon _{0}}\left [ -\frac{q^{2}}{d}+\frac{2\; q\; Q\; d}{D^{2}} \right ]

  • Option 3)

      \frac{1}{4\pi \epsilon _{0}}\left [ +\frac{q^{2}}{d}+\frac{q\; Q\; d}{D^{2}} \right ]     

  • Option 4)

    \frac{1}{4\pi \epsilon _{0}}\left [ -\frac{q^{2}}{d}-\frac{q\; Q\; d}{D^{2}} \right ]

 
Answers (1)
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u=\frac{-kq^{2}}{d}+\frac{k\: q\:Q}{D+\frac{d}{2}}-\frac{k\: q\: Q}{D-\frac{d}{2}}

u=\frac{-kq^{2}}{a}+\frac{kqQ\left ( -d \right )}{D^{2}-\frac{a^{2}}{4}}

u=k\left [ \frac{-q^{2}}{a}-\frac{q\; Q\; d}{D^{2}} \right ]

u=\; \frac{1}{4\pi \epsilon _{0}}\left [ \frac{-q^{2}}{d}-\frac{q\; Q\; d}{D^{2}} \right ]

      D> > a

 

 

 

 


Option 1)

  \frac{1}{4\pi \epsilon _{0}}\left [ -\frac{q^{2}}{d}-\frac{q\; Q\; d}{2D^{2}} \right ]

Option 2)

   \frac{1}{4\pi \epsilon _{0}}\left [ -\frac{q^{2}}{d}+\frac{2\; q\; Q\; d}{D^{2}} \right ]

Option 3)

  \frac{1}{4\pi \epsilon _{0}}\left [ +\frac{q^{2}}{d}+\frac{q\; Q\; d}{D^{2}} \right ]     

Option 4)

\frac{1}{4\pi \epsilon _{0}}\left [ -\frac{q^{2}}{d}-\frac{q\; Q\; d}{D^{2}} \right ]

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