Two point electric charges of values q and 2q are kept at a distance d apart from each other in air. A third charge Q is to be kept along the same line in such a way that the net force acting on q and 2q is zero. Calculate the position of change Q in terms of q and d.

Solution

Let the charge Q be kept along the line joining two charges such that it is at a distance x from q and at a distance (d-x) from 2q.
For the net force on q and 2q to be 0 the system of charges should be in equilibrium.

Force acting on Q due to q= $\frac{1}{4 {πε}_{\circ}} \frac{q Q}{x^{2}}$
Force acting on Q du to 2q= $\frac{1}{4 {πε}_{\circ}} \frac{2 q Q}{(d - x)^{2}}$
For, the system to be in equilibrium

$\frac{1}{4 {πε}_{\circ}} \frac{q Q}{x^{2}} = \frac{1}{4 π ε_{\circ}} \frac{2 q Q}{(d - x)^{2}} \Rightarrow \frac{1}{x^{2}} = \frac{2}{(d - x)^{2}} \Rightarrow 2 x^{2} = (d - x)^{2} \Rightarrow \sqrt{2} x = (d - x) \Rightarrow x = \frac{d}{(\sqrt{2} + 1)}$

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Electric Charges And Fields

Two point electric charges of values q and 2q are kept at a distance d apart from each other in air. A third charge Q is to be kept along the same line in such a way that the net force acting on q and 2q is zero. Calculate the position of change Q in terms of q and d.

Some More Questions From Electric Charges and Fields Chapter

Figure shows tracks of three charged particles in a uniform electrostatic field. Give the signs of the three charges. Which particle has the highest charge to mass ratio?

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