Question

Two charges, +6.00 μC (left) and -2.50 μC (right) are fixed in place and separated by 1.00 m. Where (in m) along the line joining and going through the two charges will the net electric field be zero? Locate this point relative to the positive charge.

Answer 1

In electrostatic, the electric field due to point charge Q is given by

                       

  E=$\frac{KQ}{r^2}$       where K is coulomb constant = $9\times10^9$ N-m^2/C^2

According to the question

                    Electric charge $Q_{1}$ =+6.00$\mu$C (Left) and $Q_{2}$ = - 2.50 $\mu$C(Right) located by joining the line distance 1.00 m. Now let point “o” is located X m distance relative to positive charge (+6.00$\mu$C) where the net electric field is zero and remaining distance become (1-X).

$\frac{KQ_{1}}{X^2} + \frac{KQ}{(1-X)^2} =0$

$\frac{9\times10^9\times6.00\times10^{-6}}{X^2} + \frac{9\times10^9\times(-2.50)\times10^{-6}}{(1-X)^2} = 0$

after solving

$\Longrightarrow$ $\frac{6.00}{X^2} =\frac{2.50}{(1-X)^2}$

$\Longrightarrow$ $60\times(1-X)^2 = 25\times X^2$

$\Longrightarrow$ $60(1+X^2-2\times X) = 25\times X^2$

$\Longrightarrow$ $60+60\times X^2-120\times X = 25\times X^2$

$\Longrightarrow$ $35\times X^2-120\times X+60=0$

Divided this whole equation by 5 ,we get as follow

$\Longrightarrow$ $7\times X^2-24\times X+12 = 0$

 This is a quadratic equation in X. if quadratic equation is $ax^2+bx+c=0$ then its roots are found by using shridhar acharya formula as below

$x = \frac{(-b)±\sqrt{b^2-4\times a\times c}}{2\times a}$

therefore roots for above equation

$X= \frac{(-(-24))±\sqrt{24^2-4\times7\times12}}{2\times7}$

${X =\frac{(24±15.49)}{14}}$

By using positive sign

              $X=2.82 \space m$

This is an invalid answer because this distance is out of line joining of charges.

By using negative sign

          $\colorbox{aqua}{X =0.61\space m}$

This is valid answer because this is along the line joining the charges.

Therefore the distance relative to positive charge (+6.00$\mu$C) is 0.61 m.