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Question

Four positive charges $$(2\sqrt 2 - 1)$$ Q are arranged at the four corner of a square. Another charge q is placed at the centre of the square. Resulting force acting on each corner charge is zero if q is :


A
7Q4
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B
4Q7
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C
Q
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D
(2+1)Q
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Solution

The correct option is A $$\displaystyle \dfrac{-7Q}{4}$$
The total force at any of the corner is $$F=\sqrt2k(2\sqrt2-1)^2Q^2/a^2+k(2\sqrt2-1)^2Q^2/(a\sqrt2)^2+k(2\sqrt2-1)Qq/(\dfrac{\sqrt2 a}{2})^2$$ where $$k=1/4\pi\epsilon_o$$ 
$$\Rightarrow F=k\dfrac{(2\sqrt2-1)^2(2\sqrt2+1)Q}{2a^2}+k\dfrac{4(2\sqrt2-1)q}{2a^2}=0\Rightarrow q=-(2\sqrt2+1)(2\sqrt2-1)Q/4$$
$$\Rightarrow q=-7Q/4$$

Physics

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