Question

+4q and -q are two charges separated by a distance 'd’ metre. Where should be a third charge +Q be placed to attain equilibrium?

Answer 1

Given:

+4q and -q are two charges separated by a distance of "d" metre.

To find:

Position where a third charge +Q can be placed to attain equilibrium.

Concept:

We have to place the charge +Q nearer to the -q charge (as compared to +4q charge) because it has lower magnitude.

In order to attain equilibrium , +Q charge will be attracted by -q charge and will be equally repelled by +4q charge.

Refer to the diagram to understand the location of the charges.

Calculation:

Let +Q be located x metres from -q charge , so it will be located (x + d) m from +4q charge. The forces will be equal and opposite.

$\therefore \: \dfrac{1}{4\pi\epsilon_{0}} \bigg \{ \dfrac{(4q)(Q)}{ {(d + x)}^{2} } \bigg \} = \dfrac{1}{4\pi\epsilon_{0}} \bigg \{ \dfrac{(q)(Q)}{ {x}^{2} } \bigg \}$

Cancelling the common terms:

$= > \dfrac{(4q)(Q)}{ {(d + x)}^{2} } = \dfrac{(q)(Q)}{ {x}^{2} }$

Cancelling the charge factors:

$= > \dfrac{4}{ {(d + x)}^{2} } = \dfrac{1}{ {x}^{2} }$

Taking square root on both sides:

$= > \dfrac{2}{ d + x } = \dfrac{1}{ x }$

$= > 2x = d + x$

$= > 2x - x = d$

$= > x = d$

So , +Q charge will be located d metres from -q charge and (d+d) = 2d from +4q charge.