Question

Two positive charge is of +4 microcoulomb & 16 micro coulomb are seperated by distance of 20 cm. Where how much magnitude & of what nature of charge to be placed on the line joining the 1st two in between them so as all of 3 charges remain in equilibrium

Answer 1

Given :

Magnitude of 1st charge = $+ 4 \mu C$

Magnitude of 2nd charge = $+ 16 \mu C$

Distance between these two charges = 20 cm

To Find :

Position , magnitude and nature of the third charge to be placed on the line joining the 1st two in between them so as all of 3 charges remain in equilibrium = ?

Solution :

For all the charges to be in equilibrium , the third charge must be negative .

Let a charge $-q \mu C$ be placed at a distance x from the   $+ 4 \mu C$ charge .

So , now net force acting on -q is :

=$\frac{Kq\times 16}{(20-x)^2} - \frac{Kq\times 4}{x^2}$

This force must be equal to zero for equilibrium , so :

$\frac{Kq\times 16}{(20-x)^2} - \frac{Kq\times 4}{x^2}= 0$            -(1)

Net force acting on   $+ 4 \mu C$ charge should also be zero , so :

$\frac{-K\times 4 \times 16}{(20)^2} + \frac{Kq\times 4}{x^2}= 0$          -(2)

Net force on $+ 16 \mu C$ charge also be zero , so :

$\frac{K\times 4 \times 16}{(20)^2} + \frac{Kq\times 16}{(20-x)^2}= 0$         - (3)

From eq (1) :

$\frac{16Kq}{(20-x)^2} = \frac{4Kq}{x^2}$

Or, $\frac{4}{(20-x)^2} = \frac{1}{x^2}$

Or, $4x^2 = 400 = x^2 - 40x$

Or, $3x^2 + 40x- 400 = 0$

So , $x = \frac{-40 \pm \sqrt{1600 +4800} }{6}$ =$\frac{-40 \pm 80 }{6} = \frac{40}{6}$

So , x = 6.6  cm

Now putting x = 6.6 in eq (2) :

$\frac{4Kq}{(6.6)^2}= \frac{64K}{20^2}$

Or, $\frac{q}{(6.6)^2} = \frac{16}{400}$

So, $q = \frac{16}{400}\times (6.6)^2$

        = $1.74 \mu C$

Hence , the required charge is negative with a magnitude of  $1.74 \mu C$ and placed at a distance of 6.6 cm from $+4\mu C$ charge .