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# Engineering

Four point charges -q,+q \:,\:-q\ and \ +q are place on y-axis at  y=-2d,y=-d,y=+d\:\:and\:\:y=+2d respectively . The magnitude of the electric field E at a point on the x-axis at x=D, with D\gg d, will behave as :

 

  • Option 1)

    E\propto \frac{1}{D^{3}}
     

  • Option 2)

    E\propto \frac{1}{D}

  • Option 3)

    E\propto \frac{1}{D^{4}}

     

  • Option 4)

    E\propto \frac{1}{D^{2}}

 

Answers (1)
S solutionqc

E=\frac{2kqD}{(D^{2}+d^{2})^{3/2}}-\frac{2kqD}{(D^{2}+4d^{2})^{3/2}}

\Rightarrow E= 2kq D \left [ (D^{2}+d^{2})^{-3/2} -(D^{2}+4d^{2})^{-3/2}\right ]

\Rightarrow E= \frac{2kq D}{D^{3}} \left [ \left ( 1+\frac{d^{2}}{D^{2}} \right )^{-3/2}-\left ( 1+\frac{4d^{2}}{D^{2}} \right )^{-3/2}\right ]    

\Rightarrow E= \frac{2kq }{D^{2}} \left [ \left ( 1+\frac{3}{2}\frac{d^{2}}{D^{2}} \right )-\left ( 1-\frac{3}{2}\frac{4d^{2}}{D^{2}} \right )\right ]

Applying binomial approximation \because d < < < D

E=\frac{9kq\:d^{2}}{D^{4}}

\Rightarrow E \propto \frac{1}{D^{4}} 

 


Option 1)

E\propto \frac{1}{D^{3}}
 

Option 2)

E\propto \frac{1}{D}

Option 3)

E\propto \frac{1}{D^{4}}

 

Option 4)

E\propto \frac{1}{D^{2}}

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