Four charges equal to   -Q  are placed at the four corners of a square and a charge q   is at its centre If the system is in equilibrium the value of q is 

  • Option 1)

    -\frac{Q}{4}\left ( 1+2\sqrt{2} \right )

  • Option 2)

    \frac{Q}{4}\left ( 1+2\sqrt{2} \right )

  • Option 3)

    -\frac{Q}{2}\left ( 1+2\sqrt{2} \right )

  • Option 4)

    \frac{Q}{2}\left ( 1+2\sqrt{2} \right )

 

Answers (2)

As we learnt in

Superposition of Electric field -

The resultant electric field at any point is equal to the vector sum of all the electric fields.

 

- wherein

\vec{E}=\vec{E_{1}}+\vec{E_{2}}+\vec{E_{3}}+\cdots\vec{E_{n}}

 

 Consider the four forces F1, F2, F3 and F4  acting on charge (-Q)   placed at A

Distance \: \: CA= \sqrt{2}a

Distance \: \:EA= \frac{\sqrt{2}a}{2}= \frac{2}{\sqrt{2}}

For equilibrium, consider forces along DA and  equate the resultant to zero

\therefore \frac{1}{4\pi \varepsilon _{0}}\frac{Q\times Q}{\left ( DA \right )^{2}}+ \frac{1}{4\pi \varepsilon _{0}}\frac{Q\times Q}{\left ( CA \right )^{2}}\cos 45^{0}

- \frac{1}{4\pi \varepsilon _{0}}\frac{Q\times q}{\left ( EA \right )^{2}}\cos 45^{0}= 0

or \: \: \: \: \: \frac{Q}{a^{2}}+\frac{Q}{2a^{2}}\times \frac{1}{\sqrt{2}}-\frac{q}{a^{2}/2}\times \frac{1}{\sqrt{2}}=0

or \: \: \: \: \: Q\left [ 1+\frac{1}{2\sqrt{2}} \right ]= q\sqrt{2}

or \: \: \: \: \:q= \frac{Q}{\sqrt{2}}\left [ \frac{2\sqrt{2}+1}{2\sqrt{2}} \right ]=\frac{Q}{4}\left ( 1+2\sqrt{2} \right )

 


Option 1)

-\frac{Q}{4}\left ( 1+2\sqrt{2} \right )

Incorrect

Option 2)

\frac{Q}{4}\left ( 1+2\sqrt{2} \right )

Correct

Option 3)

-\frac{Q}{2}\left ( 1+2\sqrt{2} \right )

Incorrect

Option 4)

\frac{Q}{2}\left ( 1+2\sqrt{2} \right )

Incorrect

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