Ncert Exercises & Solutions

Two charges q and -3q are placed fixed on x-axis separated by distance d. Where should a third charge 2q be placed such that it will not experience any force?

1.	At a distance \(\dfrac{d}{\sqrt3 -1}\) to the left of charge \(q\)
2.	At a distance \(\dfrac{d(\sqrt3 -1)}2\) to the left of charge \(q\)
3.	At a distance \(\dfrac{d}{2}\) to the right of charge \(-3q\)
4.	Midway between \(q\) and \(-3q\)

Hint: Use Coulomb's law.

Step 1: Find the location of the third charge.

Here, let us keep the charge 2q at a distance r from A.

Thus, the charge 2q will not experience any force when the force of repulsion on it due to q is balanced by the force of attraction on it due to -3q, at B, where AB = d.

Step 2: Find the net force on the charge and put it equal to zero.

Thus, force of attraction by -3q = Force of repulsion by q

\Rightarrow \frac{2 q \times q}{4 {πε}_{0} x^{2}} = \frac{2 q \times 3 q}{4 {πε}_{0} (x + d)^{2}}

\Rightarrow {(x + d)}^{2} = 3 x^{2}

\Rightarrow x^{2} + d^{2} + 2 xd = 3 x^{2}

\Rightarrow 2 xd = 2 x^{2} - d^{2}

\Rightarrow 2 x^{2} - 2 dx - d^{2} = 0

\Rightarrow x = \frac{d}{2} \pm \frac{\sqrt{3} d}{2}

Step 3: Choose the value with a positive sign.

(For negative sign, the position of the charge will be between q and -3q and hence is unadaptable.)

x = \frac{d}{2} + \frac{\sqrt{3} d}{2}

= \frac{d}{2} (1 + \sqrt{3})

to the left of q.

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