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  • Solve! Two point charges (+Q) and (– 2Q) are fixed on the X-axis at positions a and 2a from origin respectively. At what position on the axis, the resultant electric field is zero

Two point charges (+Q) and (– 2Q) are fixed on the X-axis at positions a and 2a from origin respectively. At what position on the axis, the resultant electric field is zero

  • Option 1)

    Only x = \sqrt2 a

  • Option 2)

    Only x = -\sqrt2 a

  • Option 3)

    Both x = \pm \sqrt2 a

  • Option 4)

    x = \frac{3a}{2}

 

Answers (1)

best_answer

As we have learnt,

 

Neutral Point due to two unlike Point -

\dpi{100} \frac{Q_{1}}{Q_{2}}=\left ( \frac{x}{x+l} \right )^{2}

- wherein

Where, x = distance between Q1 and Q2

Let the electric field is zero at a point P distance d from the charge +Q so at P.

\frac{kQ}{d^2} + \frac{k(-2Q)}{(a+d)^2} = 0 \\*\Rightarrow \frac{1}{d^2} = \frac{2}{(a+d)^2} \Rightarrow d =\frac{a}{\sqrt2 -1}

Since d> a i.e. point P must lies on negative x-axis as shown at a distance x from origin hence

x = d - a = \frac{a}{\sqrt2 -1} -1 = \sqrt2 a

Actually P lies on negative x-axis so x =- \sqrt2 a.


Option 1)

Only x = \sqrt2 a

Option 2)

Only x = -\sqrt2 a

Option 3)

Both x = \pm \sqrt2 a

Option 4)

x = \frac{3a}{2}

Posted by

Aadil

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