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}

 

Answers (1)
V Vakul

As we learnt in

Potential of a System of Charge -

V=frac{kq_{1}}{r_{1}}+kfrac{q_{2}}{r_{2}}+frac{kleft ( -q_{3} 
ight )}{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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