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. Three identical dipoles are arranged as shown below. What will be the net electric field at P(k=\frac{1}{4 \pi\varepsilon _0})

  • Option 1)

    \frac{kp}{x^3}

  • Option 2)

    \frac{2kp}{x^3}

  • Option 3)

    zero

  • Option 4)

    \frac{\sqrt2 kp}{x^3}

 

Answers (1)

As we have learned

At general Point -

E_{g}=\frac{1}{4\pi \epsilon _{0}}\frac{P}{r^{3}}\sqrt{3\cos ^{2}\Theta +1}

V_{g}=\frac{1}{4\pi \epsilon _{0}}\frac{P\cos \Theta }{r^{2}}

-

 

 

 Point P lies at equatorial positions of dipole 1 and 2 and axial position of dipole 3.

Hence field at P

 

due to dipole 1 

E_1= \frac{kp}{x^3}    (towards left)

due to dipole 2

E_2= \frac{kp}{x^2}    (towards left)

due to dipole 3  E_3= \frac{k(2p)}{x^3}    (towards right)

So net field at P will be zero.

 


Option 1)

\frac{kp}{x^3}

Option 2)

\frac{2kp}{x^3}

Option 3)

zero

Option 4)

\frac{\sqrt2 kp}{x^3}

Posted by

Vakul

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