Four point charges -Q, -q, 2q and 2Q are placed, one at each corner of the square. The relation between Q and q for which the potential at the centre of the square is zero is: Option 1) Q = -q Option 2) $Q=\frac{1}{q}$ Option 3) Q = q Option 4) $Q=\frac{-1}{q}$

V Vakul

As we learnt in

Potential of a System of Charge -

$\dpi{100} V=\frac{kq_{1}}{r_{1}}+k\frac{q_{2}}{r_{2}}+\frac{k\left ( -q_{3} \right )}{r_{3}}+\cdots$

- wherein

$AC=BD=\sqrt{2a}$

$OA=OB=OC=OD=\frac{a\sqrt{2}}{2}=\frac{a}{\sqrt{2}}$

Potential at the centre O due to given charge configuration is

$V=\frac{1}{4\pi \varepsilon _{o}}\:[\frac{(-Q)}{(\frac{a}{\sqrt{2}})}+\frac{(-q)}{(\frac{a}{\sqrt{2}})}+\frac{(2q)}{(\frac{a}{\sqrt{2}})}+\frac{2Q}{(\frac{a}{\sqrt{2}})}]\:=0$

-Q-q+2q+2Q=0          Q+q=0      or Q= - q

Option 1)

Q = -q

Correct

Option 2)

$Q=\frac{1}{q}$

Incorrect

Option 3)

Q = q

Incorrect

Option 4)

$Q=\frac{-1}{q}$

Incorrect

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